Height, Time of the Ball The height $h$ in feet of a baseball hit $3$ feet above the ground is given by $h=3+75t-16t²$ where $t$ is the time in seconds.
a. Find the height of the ball after two seconds
b. Find the time where the ball hits the ground in the field
c. What is the maximum height of the ball?

a. Find the height of the ball after two seconds: 
Set $t=2$, so $h=3+75t-16t=3+75(2)-16(2)^2=\color{#180}{89}$ feet.
b. Find the time where the ball hits the ground in the field: 
Solve for $t$ such that $h=0$ as the height above the ground will be zero at the point of impact, so $0=3+75t-16t² \implies t=\color{#180}{4.73}$ seconds.
c. What is the maximum height of the ball?: 
Differentiate the expression for $h$ with respect to time and set it equal to zero to find the stationary points (which are maximums or minimums) $\frac{\mathrm{d}h}{\mathrm{d}t}=75-32t=0 \implies t=2.34375$ seconds. Now substitute this value into the expression for $h$ such that $h(2.34375)=\color{#180}{90.9}$ feet, which is the maximum height. We note that the second derivative $\frac{\mathrm{d^2}h}{\mathrm{d}t^2}=-32 \lt 0$. So the stationary point is indeed a maximum (and not a minimum). 
