Question

What is the velocity at the midway point of a ball able to reach a height y when thrown with upward with an initial velocity v0?

Answer 1

Let's first find the time that the ball takes to reach the maximum height:

$$v=v_0-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}.$$

Then, we can find the maximum height reached by the ball:

$$H=v_0t-\dfrac{1}{2}gt^2,$$ $$H=v_0\dfrac{v_0}{g}-\dfrac{1}{2}g(\dfrac{v_0}{g})^2=\dfrac{v_0^2}{2g}.$$

Let's find the time that the ball takes to reach the midpoint:

$$v=v_0-gt_1,$$ $$t_1=\dfrac{v_0-v}{g}.$$

Then, we can find the midpoint height reached by the ball:

$$h=v_0t_1-\dfrac{1}{2}gt_1^2,$$ $$h=v_0\dfrac{v_0-v}{g}-\dfrac{1}{2}g(\dfrac{v_0-v}{g})^2,$$ $$h=\dfrac{v_0^2-v^2}{2g}.$$

Since $h=\dfrac{H}{2}$, we can write:

$$\dfrac{v_0^2-v^2}{2g}=\dfrac{v_0^2}{4g},$$ $$2v_0^2-2v^2=v_0^2,$$ $$v=\dfrac{v_0}{\sqrt{2}}.$$