Question A.1: A ball is dropped from the top of a building 50.0 m tall. Ho...

First, write the general ballistics equation for this situation:

x=\frac{1}{2} a t^{2}+v_{0} t+x_{0}

Here, a=-9.80 m / s ^{2} is the acceleration of gravity that causes the ball to fall, v_{0}=0 is the initial velocity, and x_{0}=50.0 m is the initial position. Substitute only the initial velocity, v_{0}=0 , to obtaining the following equation:

x=\frac{1}{2} a t^{2}+x_{0}

This equation must be solved for t. Subtract x_{0} from both sides:

x-x_{0}=\frac{1}{2} a t^{2}+x_{0}-x_{0}=\frac{1}{2} a t^{2}

Multiply both sides by 2/a:

\left(\frac{2}{a}\right)\left(x-x_{0}\right)=\left(\frac{2}{a}\right) \frac{1}{2} a t^{2}=t^{2}

It’s customary to have the desired value on the left, so switch the equation around and take the square root of both sides:

t=\pm \sqrt{\left(\frac{2}{a}\right)\left(x-x_{0}\right)}

Only the positive root makes sense. Values could now be substituted to obtain a final answer