Question

A ball is thrown straight up from the edge of a 15 foot balcony with initial velocity 40 feet/sec. how high will the ball travel?

Answer 1

Hello
The initial height of the ball is $S_i = 15~ft = 4.6~m$, while the initial velocity is $v_i = 40ft/s = 12.2~m/s$.
This is a uniformly accelerated motion, where the acceleration is the gravitational acceleration: $a=-9.8~m/s^2$, where the negative sign means that the acceleration points downwards. 
We can proceed in two steps: 

1) First of all, we calculate the time the ball needs to reach the maximum height. At the point of maximum height, the velocity is zero: $v_f = 0m/s$. 
We can use the formula
$a= \frac{v_f-v_i}{\Delta t}$
To find the time:
$\Delta t= \frac{v_f-v_i}{a} = 1.2 s$

2) At this point, we can calculate the total height reached by the ball after this time, i.e. the maximum height:
$S=S_i+v_i \Delta t+ \frac{1}{2} a (\Delta t)^2 =$
$= 4.6~m+(12.2~m/s)(1.2~s)+ \frac{1}{2}(-9.8~m/s^2)(1.2~s)^2 = 12.1~m$