Question

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 82 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to _____

Answer 1

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

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

Where:  

$y=0$ is the height of the ball when it hits the ground  

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball  

$t$ is the time when the ball strikes the ground

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

Having this clear, let's find $t$ from (1):  

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)  

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)  

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$, which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$  (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$  (5)

Solving (5) we find the positive result is:

$t=17.68 s$