Question

At the top of a 125 meter tall tower, two baseballs are released from rest. One is dropped, and the other is thrown horizontally with initial speed v. If the baseballs land 30 meters apart, then what is the initial speed v?

Answer 1

First, we need to calculate the amount of time that the balls are falling, using the formula:

$\Delta S=\frac{gt^2}{2}$

Where DeltaS is the distance traveled, g is the gravity acceleration and t is the time.

So, for DeltaS = 125 and g = 10, we have:

$\begin{gathered} 125=\frac{10t^2}{2} \\ 125=5t^2 \\ t^2=25 \\ t=5\text{ s} \end{gathered}$

Now, to find the initial speed v, we can use the formula for the distance traveled in the horizontal movement:

$\Delta S=V\cdot t$

For DeltaS = 30 and t = 5, we have:

$\begin{gathered} 30=V\cdot5 \\ V=\frac{30}{5} \\ V=6\text{ m/s} \end{gathered}$

So the initial speed v is 6 m/s.