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anonymous

  • 4 years ago

A tennis ball is thrown vertically upwards with an initial velo. of 5m/s.Assuming the acceleration due to gravity is 10m/s^2. find the max gain in height of the ball.

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  1. JamesJ
    • 4 years ago
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    When the ball is at its maximum height all of its energy is in gravitational potential energy, PE. We can express the PE of an object of mass m at a height as PE = mgh. When it thrown, all of its energy is kinetic energy, KE. What's the expression for KE? When you know it, set it equal to the expression for PE and you'll be able to solve for h, the height.

  2. anonymous
    • 4 years ago
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    but this is supposedly a kinematics question....

  3. JamesJ
    • 4 years ago
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    This is kinematics

  4. JamesJ
    • 4 years ago
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    \[ KE = \frac{1}{2}mv^2 \] Set KE = PE and you can find an expression for h ...

  5. anonymous
    • 4 years ago
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    25/20meters...in ideal conditions

  6. JamesJ
    • 4 years ago
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    If PE = KE then \[ mgh = \frac{1}{2}mv^2 \] Cancel m from both sides and divide by g, then \[ h = \frac{v^2}{2g} \] See now how to replicate antrio's answer?

  7. JamesJ
    • 4 years ago
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    (@antrio: welcome to OpenStudy! fyi, we try and avoid giving answers outright. Much better to show people how to get to the answer themselves.)

  8. anonymous
    • 4 years ago
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    wong@ the questions can be solved by 3rd eqn of motion...or by conservation of energy.both yeilds same result....james is right

  9. JamesJ
    • 4 years ago
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    Jamelynn, talk to me ... where are you stuck?

  10. anonymous
    • 4 years ago
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    okay james@

  11. anonymous
    • 4 years ago
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    i found out from my notes another way to solve the prob by using the equation for constant acceleration: s=u^2 + 2as s=5^2 + 2(-10)s 21s=25 s=1.19

  12. JamesJ
    • 4 years ago
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    The equation you've written down is not correct. What is correct is that if u is initial velocity and v final, then \[ v^2 = u^2 + 2as \] At the top of its arc, the object has no velocity, so v = 0. Hence \[ s = - \frac{u^2}{2a} \] The acceleration here is gravity, hence a = -g. Thus \[ s = \frac{u^2}{2g} \] This is identical to the equation above, become there v is also the initial velocity. In short, your current answer is wrong.

  13. anonymous
    • 4 years ago
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    there is mistake in that equ....the correct eqn is s=u x time + 1/2 g time^2..........and it cannot be applied here as time is not given

  14. anonymous
    • 4 years ago
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    thx, i realized there was a mistake in my formula

  15. JamesJ
    • 4 years ago
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    So one more time All the kinetic energy when the ball is thrown is converted at the top of its flight into gravitational potential energy KE(when ball thrown) = PE(when ball at the top of its flight) Let \( u \) be the velocity at which it is thrown. Then \[ KE = \frac{1}{2}mu^2 \] Let \( s \) be the height at the top of its path. Then \[ PE = mgs \] Hence \( KE = PE \) implies \[ \frac{1}{2}mu^2 = 2gs \] Solving for s, we find that \[ s = \frac{u^2}{2g} \] Given that \[ u = 5 \ m/s \] and \[ g = 10 \ m/s^2 \] then \[ s = \frac{u^2}{2g} = \frac{5^2}{2(10)} = \frac{25}{20} = 1.25 \ m \]

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