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    read the scenario, then answer the questions. a 2 kg ball is thrown upward with a velocity of 15 m/s. what is the kinetic energy of the ball as it is being thrown? j what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? j

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    get read the scenario, then answer the questions. a 2 kg ball is thrown upward with a velocity of 15 m/s. what is the kinetic energy of the ball as it is being thrown? j what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? j from EN Bilgi.

    A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown? J What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? J?

    Answer (1 of 2): Question: What is the kinetic energy of the ball as it is being thrown? Answer: E_K=\frac {1}{2}mv^2 E_K=\frac {1}{2}(2)(15)^2 E_K=225\ J Question: What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? Answer wi...

    A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown? J What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? J?
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    2 Answers
    Eduardo Dequilla
    , B.S. Physics from Ateneo de Davao University (2005)
    Answered 1 year ago · Author has 6.9K answers and 5.6M answer views
    Given : m = 2 kg ; v = 15 m/s ; kinetic energy, KE = ? ; potential energy , PE = ? ; g = 9.8 m / s^2 Solving for the kinetic energy, KE KE = 1/2 mv^2 KE = 0.5 * 2 kg * (15 m/s)^2 KE = 225 Joules The kinetic energy of the ball as it is being thrown is 225 Joules. What is the potential energy PE of the ball when it gets to its maximum height Solving for the potential energy PE PE = mgh PE = KE … from the law of conservation of mechanical energy. At the maximum height of the ball its KE = 0 since all become PE. PE = 225 Joules The potential energy of the ball when it gets to its maximum height just before fGiven : m = 2 kg ; v = 15 m/s ; kinetic energy, KE = ? ; potential energy , PE = ? ; g = 9.8 m / s^2 Solving for the kinetic energy, KE KE = 1/2 mv^2 KE = 0.5 * 2 kg * (15 m/s)^2 KE = 225 Joules The kinetic energy of the ball as it is being thrown is 225 Joules. What is the potential energy PE of the ball when it gets to its maximum height Solving for the potential energy PE PE = mgh PE = KE … from the law of conservation of mechanical energy. At the maximum height of the ball its KE = 0 since all become PE. PE = 225 Joules The potential energy of the ball when it gets to its maximum height just before falling back to the ground is 225 J. By using PE = mgh PE = 2 kg * 9.8 m/s^2 * h ….. the value of h needs to be solved first using v^2 = 2gh Solving for h h = v^2 / 2g h = 225 / 19.6 h = 11.48 m PE = 2 kg * 9.8 m/s^2 * 11.48 m PE = 225 J The potential energy of the ball when it gets to its maximum height just before falling back to the ground is 225 J.
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    A 2kg ball is thrown upward with a velocity of 15 MS. What is the kinetic energy of the ball as it is being thrown?How high should a 2 kg mass be lifted from the ground if it is thrown upward at a speed of 15 m/s assuming gravity=10m/s²?A 1.5kg object is thrown horizontally with an initial velocity of 10m/s from a height of 11m. What will be the kinetic energy of the object just before it hits the ground?A body of mass 5 kg is thrown in the upward direction with a velocity of 50 m/s. At what height will the potential and kinetic energy be the same?A 0.15kg ball is thrown upward. If the gravitational potential energy of the ball on its highest position is 1.80J relative to its initial point, what height did the ball reach?
    Zahidullah Noori
    , Online Maths and Physics Tutor
    Answered 1 year ago
    Question:What is the kinetic energy of the ball as it is being thrown?

    Answer:EK=12mv2EK=12mv2

    EK=12(2)(15)2EK=12(2)(15)2 EK=225JEK=225J Question:What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?

    Answer without any calculation:Assuming that the ball has not gained or lost energy from or to any other sources, the potential energy at the highest point should be equal to the initial kinetic energy. Hence, the answer is 225 J

    Answer with the calculations:

    First, find the maximum height:Hint: the velocity at the maximum point will be

    Question:What is the kinetic energy of the ball as it is being thrown?

    Answer:EK=12mv2EK=12mv2

    EK=12(2)(15)2EK=12(2)(15)2 EK=225JEK=225J Question:What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?

    Answer without any calculation:Assuming that the ball has not gained or lost energy from or to any other sources, the potential energy at the highest point should be equal to the initial kinetic energy. Hence, the answer is 225 J

    Answer with the calculations:

    First, find the maximum height:Hint: the velocity at the maximum point will be zero.2gs=v2u22gs=v2u2

    2(9.807)s=(0)2(15)22(9.807)s=(0)2(15)2 s=22519.614s=22519.614 s=11.47ms=11.47m

    Use the formula for potential energy:EP=mghEP=mgh

    EP=(2)(9.807)(11.47)EP=(2)(9.807)(11.47) EP=224.97JEP=224.97J The reason we get just a little bit of difference in calculation version and without calculation version is because of rounding off of values. In case you don’t understand the formulas used, either ask in comment or ask another question specifically for that purpose.
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    Ashish, Canadian
    Answered Jan 17, 2022 · Upvoted by Gene Ewald, M.S. Physics, University of Minnesota - Twin Cities (1971) · Author has 687 answers and 125.4K answer views
    A body of mass 5 kg is thrown in the upward direction with a velocity of 50 m/s. At what height will the potential and kinetic energy be the same?

    Lets say at height h meters, we have PE = KE

    mgh = ½m𝑣²

    h =v²2g𝑣²2g ...(1)

    If initial velocity is 𝑢 m/s, then at height h, the velocity 𝑣 will be given by

    𝑣² = 𝑢² - 2gh …..(2) Plug (2) in (1)

    h =u²2gh2g𝑢²2gh2g

    2gh = 𝑢² - 2gh 4gh = 𝑢²

    h = u²4g𝑢²4g

    h = (50)²49.8(50)²49.8

    h ≈ 63.8 m
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    Anwar Osman, knows ArabicAnswered Nov 14, 2021
    How high should a 2 kg mass be lifted from the ground if it is thrown upward at a speed of 15 m/s assuming gravity=10m/s²?since,PE=mgh and KE=0 ,ON THE GROUND and KE=0.5mv^2 and PE=0, on top,then KE =0.5*2kg*(15 M/s)^2=>112.5kg m^2/s^2 at the highest height and PE =2kg*10 m/s^2*h=>20kg m/s^2*h. at the ground back ,so rising and falling height will be doubled the height so dividing KE to PE multiplied by 2=>11.25m
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    Robert Toop
    Answered 7 months ago · Author has 2.9K answers and 707.4K answer views
    A ball is thrown vertically upward and returns to its original position in 15 seconds. What is the initial velocity and the maximum height reached by the ball?Total time = 15 s. Without air resistance, rise time = fall time = 15/2 s Kinematics equation for velocity under constant acceleration a, as a function of time t: v = u + at ———— [1] Initial speed = u Final speed v = -u back at initial height a = -9.81 m/s^2 due to gravity Substitute our values into equation 1: -u = u -9.81*15 -2u = -9.81*15 u = 73.6 m/s up Since speed during rise changes from u to 0, average speed = u/2 Max height = (average speed) * (rise time) = u/2*7.5 = 276 m
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    Arega Tayu
    Answered 1 year ago · Author has 1.5K answers and 308.2K answer views
    A ball of mass 1.5kg thrown vertically upward with a speed of 10m/s from the top of a tree 10m high. What is the total energy waisted before it hits the ground?
    To start with energy cannot be wasted. “Energy can neither be created nor destroyed” (low of conservation of energy). Energy can be changed from one form to another now let’s continue. On its way down reaching the top of the tree, velocity of the ball =Vi=20m/s the tree is 10 m heigh => the energy stored in the ball at that point is part Kinetic and part potential. E = 1/2 mv^2 + mgh = m(v^2/2 + gh) = 1.5kg×(0.5(10 m/s)^2 + 9.8×10) =222 J as the ball keeps falling it is losing height. As the height aproches zero the KE approaches 222 J and PE approches zero. So thenergy before it hits the ground isTo start with energy cannot be wasted. “Energy can neither be created nor destroyed” (low of conservation of energy). Energy can be changed from one form to another now let’s continue. On its way down reaching the top of the tree, velocity of the ball =Vi=20m/s the tree is 10 m heigh => the energy stored in the ball at that point is part Kinetic and part potential. E = 1/2 mv^2 + mgh = m(v^2/2 + gh) = 1.5kg×(0.5(10 m/s)^2 + 9.8×10) =222 J as the ball keeps falling it is losing height. As the height aproches zero the KE approaches 222 J and PE approches zero. So thenergy before it hits the ground is 222 J Or max height=Vi^2/(2×9.8)=100/19.6 m => PE at the max height = mgh =1.5×9.8×(100/19.6) = 222 J PE at the top = KE at the bottom =>energy before it touch the ground is 222 J.
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    Leslie Schultz, Jump Gate Supervisor at Graviton Industries (1983-present)
    Answered 1 year ago · Author has 1.3K answers and 373.1K answer views
    What is the ratio of potential energy to kinetic energy 5m above the ground when falling?If the object is dropped with initial velocity of 0 m/s from a height of 5 m - the ratio of potential to kinetic energy equals infinity. The potential energy can be defined as Ep = mgh, where h=5m. Kinetic energy equals Ek=0, as the initial velocity at h=5m equals zero. This is abadly formulated question, BTW.
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    Michael Packer, former Process Engineer
    Answered 2 years ago · Author has 517 answers and 399.2K answer views
    A tennis ball goes vertically upwards with a speed of 15 m/s to reach the maximum height b. What is this height?Use Equation of Motion in vertical or y-direction only max height = b = v0y*t - 0.5*g*t^2 where: g = 9.81 m/s^2 v0y = 15 m/s and at max height, db/dt = v0y - g*t = 0 (db/dt = 0) and solve for t, time to reach max height, b: t = v0y/g = (15 m/s)/(9.81 m/s^2) = 1.53 seconds to reach max height b = (15 m/s)*(1.53 s) - 0.5*(9.81 m/s^2)*(1.53.s)^2 = 11.47 m ~ 11.5 m max height of tennis ball
    298 views · Answer requested by Muntadher Ahmed
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    Fiona Flora
    , works at Self Employed Lawyer
    Answered 2 years ago · Author has 95 answers and 56.2K answer views
    A 1.5kg object is thrown horizontally with an initial velocity of 10m/s from a height of 11m. What will be the kinetic energy of the object just before it hits the ground?
    Oooh, this is a cool problem! I’ll game plan it out: Kinetic energy = 1/2 m v^2 . You already know the mass, but you need to figure the velocity in order to get the joules. Velocity has 2 components here: horizontal and vertical. The horizontal component is known: 10 m/s. The vertical is gravity-induced. Falling distance = 1/2 a t^2 . Assuming Earth and g of whatever-constant-you-use, use the distance formula to determine time of falling. You know the distance and the acceleration, so solve for t. This then lets you figure the vertical component of the object’s speed, using v = a t. So, now thatOooh, this is a cool problem! I’ll game plan it out: Kinetic energy = 1/2 m v^2 . You already know the mass, but you need to figure the velocity in order to get the joules. Velocity has 2 components here: horizontal and vertical. The horizontal component is known: 10 m/s. The vertical is gravity-induced. Falling distance = 1/2 a t^2 . Assuming Earth and g of whatever-constant-you-use, use the distance formula to determine time of falling. You know the distance and the acceleration, so solve for t. This then lets you figure the vertical component of the object’s speed, using v = a t. So, now that you know both the vertical and horizontal components, you can use the Pythagorean theorem (a^2 + b^2 = c^2) to compute the combined speed of the object. In the equation above, that is “c”. You don’t actually have to get to c, but just c^2, which means you just have to add the two components together in order to get the “v^2” portion of the kinetic energy formula at the top. Here, a and b are the horizontal and vertical velocities. Cake!
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    Ankit Gautam, Centre Forward at FC Barcelona
    Answered 3 years ago · Author has 110 answers and 59.3K answer views
    A 500 g stone is thrown up with a velocity of 15 m/s. What will be its potential energy at its maximum height?Thrown up Final velocity(v)=0 Initial velocity(u)=15 m/s acceleration due to gravity(g)=- 10 m/s^2 Now, v^2=u^2+2gs 0=225–20s s=11.25 m Now PE=mgh 0.5×11.25×-10 =56.25 J
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    Brad Heers
    , MSME in Applied Mechanics
    Answered 4 years ago · Author has 1.4K answers and 3.5M answer views
    A boy throws a ball nearly vertically upward from a point near the cornice of a building. It just misses the cornice on the way down and hits the ground 75 m below 5 seconds after it leaves the boy's hand. With what velocity was the ball thrown?
    The other answers (so far) are incorrect, because they fail to recognize that this problem as-defined has two unknown variables: The initial velocity AND the initial height. Implicit in the other answers is the assumption that the ball is being thrown “up” at the cornice (h=75m), but the problem as-written seems to explicitly clarify that the ball hits its peak height AT the cornice (h=75m), after being thrown upwards from a point below the cornice. If the problem is intending to have the ball thrown from the cornice itself, then it is poorly-worded. I believe that what I have described is theThe other answers (so far) are incorrect, because they fail to recognize that this problem as-defined has two unknown variables: The initial velocity AND the initial height. Implicit in the other answers is the assumption that the ball is being thrown “up” at the cornice (h=75m), but the problem as-written seems to explicitly clarify that the ball hits its peak height AT the cornice (h=75m), after being thrown upwards from a point below the cornice. If the problem is intending to have the ball thrown from the cornice itself, then it is poorly-worded. I believe that what I have described is the intended way to read the problem, and am answering it from that perspective. Go back to what you know (hopefully) about equations of motion due to gravity. We have two key equations here: (1) u(t) = u0 + v0 * t + 1/2 a * t^2 (2) v(t) = v0 + a * t where u0 and v0 are the displacement and velocity at time = 0 For purposes of consistency, let’s use “ground” as our zero point, and use positive to indicate “upwards of the ground”. What do we know? I: Total “flight time” is 5 seconds. II: Acceleration is -9.81 m/sec^2 III: At t = 5 seconds, u(5) = 0 IV: At t = 0 seconds, v0 is “positive” V: At u(t) = 75m, we know that v(t) = 0 (since v(t) is the top height, and at the top height the velocity is zero. And the problem notes that this happens right at the cornice, which is at 75m. Let’s call this point in time “t1”. Thus v(t1)=0 and u(t1)=75m. Using the information of V, we can define a time “t2” that we can use to solve for the time it takes for the object to fall from 75m to earth (i.e. “time to get from t1 to t=5.0). We can use equation (1) with defined u1=75 and v1=0 to arrive at the time “t2”, which is the delta between five seconds and when the object reaches its peak. With this time “t2” solved for, we now know the actual time of the peak (5.0 - t2), giving us t1—this is the actual time from when the ball is thrown to when it hits its peak height of 75m. We can then re-form equation (2) and solve for v0, subject to the known t1 and v(t1). This v0 is the initial velocity. This can then be confirmed by solving for u0. Then one can again confirm that u(t1) = 75m to be sure that one has arrived at a proper answer. I have intentionally avoided writing out the equations and solving for you—I have described it in detail so that you can arrive at the answer yourself. I prefer to lead you to the answer and teach, rather than give it to you outright.
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    Read the scenario, then answer the questions. A 2 kg ball is thrown upward with a velocity

    Read the scenario, then answer the questions. A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown? ____J What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? ____J

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    Read the scenario, then answer the questions. A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown? ____J What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground? ____J

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