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.

Question

jamesj · Answer

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.

anonymous · Answer

but this is supposedly a kinematics question....

jamesj · Answer

This is kinematics

jamesj · Answer

$$ KE = \frac{1}{2}mv^2 $$

Set KE = PE and you can find an expression for h ...

anonymous · Answer

25/20meters...in ideal conditions

jamesj · Answer

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?

jamesj · Answer

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

anonymous · Answer

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

jamesj · Answer

Jamelynn, talk to me ... where are you stuck?

anonymous · Answer

okay james@

anonymous · Answer

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

jamesj · Answer

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.

anonymous · Answer

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

anonymous · Answer

thx, i realized there was a mistake in my formula

jamesj · Answer

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
$$ s = \frac{u^2}{2g} = \frac{5^2}{2(10)} = \frac{25}{20} = 1.25 \ m $$