How to find the height which a ball will bounce after a collision with ground if the upward force is known?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

From a height of $5\,m$ with respect to the ground a sphere is released. The time elapsed in the contact with the ground is $1\,ms$ and magnitude of the average upward force is $1900\,N$. Find the height (measured in meters) from the ground which the ball will bounce. Assume $g=10\,\frac{m}{s^2}$.

The alternatives are as follows:

$\begin{array}{ll}
1.&5.06\,m\\
2.&4.05\,m\\
3.&3.04\,m\\
4.&2.03\,m\\
\end{array}$

I'm confused exactly how to proceed with this question. The reason of the confusion is how to assess the answer. What I've attempted to do was to use the impulse momentum equation:

$J=\Delta p$

And the relationship between the impulse and force as follows:

$J=\overline{F}\Delta t$

In order to get the height which the sphere will get in the bounce back after collisioning with the ground I'm using the conservation of mechanical energy as follows:

$mgh=\frac{1}{2}mv^2$

Therefore:

$h=\frac{1}{2g}v^2$

but that v will be the speed attained by the ball after the collision. In order to find that v. I'm using the upward Force as follows:

At first all potential energy is transformed into kinetic energy.

$\frac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}=\sqrt{2\times 10 \times 5}= 10 \frac{m}{s^2}$

Then:

$m\delta v = \overline{F} \Delta t$

$0.1(v_f-10)=1900(10^{-3})$

$v_{f}=29$

This will be the speed attained by the ball in the bouncing back.

Then this can be used to find the height attained by the ball:

Therefore returning to the earlier equation:

$h=\frac{1}{2g}v^2$

$h=\frac{1}{2\times 10}(29)^2$

$h=42.05\,m$

But this doesn't seem very reasonable. Does it exist an error in my approach?. Did I incurred in a contradiction or something, can somebody help me here?
 
Jun 2017
337
6
Lima, Peru
I forgot to put the mass of the sphere but in this problem the mass of the sphere is $0.1\,kg$.
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
the mass hits the ground with velocity, $v=-\sqrt{2gh_0}$, which is the initial velocity for the collision.

$v_f$ is the post-collision velocity.

$F \Delta t = m[v_f-(-\sqrt{2gh_0}] \implies v_f = \dfrac{F \Delta t}{m} - \sqrt{2gh_0} = 9 \text{ m/s}$

$h_f = \dfrac{9^2}{20} = 4.05 \text{ m}$