Question

A stone is dropped from a 75-m- high building . When this stone has dropped 15 m, a second stone is thrown downward with an initial velocity such that the two stones hit the ground at the same time. What was the initial velocity of the second stone?

Answer 1

Answer: -17.66 m/s or 17.66 m/s downwards

Explanation:

The decribed situation is related to vertical motion and the main equation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)  

Where:  

$y=0 m$ is the final height of both stones

$y_{o}=75 m$ is the initial height of both stone s

$V_{o}$ is the initial velocity of the stone  

In the case of the first stone is $V_{o1}=0 m/s$ and for the second stone $V_{o2}$ is different from zero

$t$ is the time, which is the same for both stones

$g=-9.8 m/s^{2}$ is the acceleration due to gravity  

Having this clear, let's find $t$ for the first stone:

$y=y_{o}+V_{o1}t-\frac{1}{2}gt^{2}$ (2)  

$0=y_{o}+0-\frac{1}{2}gt^{2}$ (3)  

Isolating $t$:

$t=\sqrt{\frac{-2y_{o}}{g}}$ (4)  

$t=\sqrt{\frac{-2(75 m)}{-9.8 m/s^{2}}}$ (5)  

$t=3.912 s$ (6)  

Now we have to use this time to find $V_{o2}$:

$0=y_{o}+V_{o2}t-\frac{1}{2}gt^{2}$ (7)  

Isolating $V_{o2}$:

$V_{o2}=-\frac{y_{o}}{t} - \frac{gt}{2}$ (8)  

$V_{o2}=-\frac{75 m}{3.912 s} - \frac{(-9.8 m/s^{2})(3.912)}{2}$ (9)

Finally:

$V_{o2}=-17.633 m/s$ (10)  The negative sign indicates the direction of the initial velocity is vertically down