# two balls thrown upward at the same time

Ball A is thrown upward at the same time as ball B and with half the speed of ball B. a) Will ball A or B hit the ground faster? b) How much higher does B go than A?

This is one of my homework problems. But I'm confused because if the speed of ball B is twice that of ball A's, shouldn't it hit the ground first? However, the answer is that A will hit the ground before B. Why would that be so? I tried to use the displacement formula (for both a) and b)), but I'm not sure if I'm doing it right. I got the equations $x_A = 0 + v_0t - 4.9t^2$ and $x_b = 0 + 2v_0t - 4.9t^2$, but have no idea how to carry on. Also, I feel like I can get b) from the above equations, but how can I do that?

## 3 Answers

[A] Since B has more speed, it will attain more height than A. Also the time taken by it to attain velocity zero while going upward is more than that taken by A.

[B] The height attained by B is 4 times the height attained by A. The solution is as given in the picture here.

If ball A is thrown up with half the speed of B, it will not go as high, therefore will fall to the ground quicker. Using the suvat equation $s=v_{0}t+\frac{1}{2}at^2$

For A, calculate when $s=0$ to find when the ball reached the ground $$0=v_{0}t-4.5t_A^2$$ Solve for $t_A$ to get $t_A=0$ and $t_A=\frac{v_{0}}{4.5}$

Do the same for B: $$0=2v_{0}t_B-4.5t_B^2$$ $t_B=0$ and $t_B=\frac{2v_{0}}{4.5}$

This shows that it takes twice as long for B to reach the ground.

To find the heights reached by each ball use the equation $v^2=v_0^2+2as$. For A this will give: $$0=v_0^2-19.6s_A$$ $$s_A=\frac{v_0^2}{19.6}$$ The same can be done for B to give $s_B=\frac{4v_0^2}{19.6}$.This shows that B will reach a height 4 times greater than A.

When each ball reaches the top of its flight, it momentarily stands still and then starts falling. It doesn't really matter what happens before this point, it just starts falling now.

• Ball B surely reaches higher than ball A. If they reach their tops simultaneously, then they would starts their free falls at the same time and A would hit the ground first because it starts closer to the ground. This argument alone means that A wins this race.
• At the same time A reaches its top before B does. B must be slowed down from a larger speed than A before reaching its top. That takes a longer time. A therefore reaches its top and starts it free fall before B does - another argument for A winning this race.

Regarding the math, you are on the right track with your two equations. Just include the fact that the end-positions will be zero as well, $x_A=0$ and $x_B=0$. Now you have two equations with two unknowns and can solve it.

To solve question b), I'll give a hint: Look for the "signal". You are looking for the top of their flights. The "signal" there is - as mentioned - that the speed is zero. Now, find a fitting equation that involves both the speed and distance (but not time, since we do not know the time to reach that top) and solve for distance.