Answer
Taking $g=-32.17feet/s^2$.
(a) Time at which tennis ball is at maximum height is $t=\frac{32g+5}{8g}=3.981s$
(b) Maximum height reached by the tennis ball is $y=5-\frac{(32g+5)^2}{128g}=259.866feet$
Work Step by Step
Differential equation governing the motion of the tennis ball is given as $\frac{d^2y}{dt^2}=g$.
Integrate the equation once to get
$\frac{dy}{dt}=gt+c_1$, where $c_1$ is a constant of integration.
Integrate it once more to get
$y=\frac{g}{2}t^2+c_1t+c_2$, where $c_2$ is also a constant of integration.
Now use the initial conditions given in the problem:
(i) At $t=0$ the ball was at the boy's head, i.e., $y=5$. Substitute that in the equation above to get
$5=\frac{g}{2}(0^2)+c_1(0)+c_2$
$c_2=5$.
(ii) At $t=8$ the ball hits the ground, i.e., $y=0$. Substitute that in the equation to get
$0=\frac{g}{2}(8^2)+c_1(8)+5$
$0=32g+8c_1+5$
$c_1=-\frac{32g+5}{8}$.
So now we get our equation of motion as
$y=\left(\frac{g}{2}\right)t^2-\left(\frac{32g+5}{8}\right)t+5$
(a) Maximum height of the motion is achieved when the first derivative of the height $y$ is zero. So we write
$\frac{dy}{dt}=gt-\left(\frac{32g+5}{8}\right)=0$
$t=\frac{32g+5}{8g}$.
This is the time when the tennis ball reaches maximum height.
(b) The value of maximum height itself can be determined by substituting the time at which the tennis ball reaches maximum height in the equation of motion the ball. So we write
$y=\left(\frac{g}{2}\right)\left(\frac{32g+5}{8g}\right)^2-\left(\frac{32g+5}{8}\right)\left(\frac{32g+5}{8g}\right)+5$
$y=\left(\frac{g}{2}\right)\left(\frac{32g+5}{8g}\right)^2-\left(\frac{g}{1}\right)\left(\frac{32g+5}{8g}\right)^2+5$
$y=-\left(\frac{g}{2}\right)\left(\frac{32g+5}{8g}\right)^2+5$
$y=5-\frac{(32g+5)^2}{128g}$.
This is the maximum height reached by the tennis ball.
