The ball height is 0, when it hits the ground. So
[tex]\begin{gathered} 0=-16t^2+96t \\ -16t(t-6)=0 \end{gathered}[/tex]
So from the equation, -16t = 0 or (t - 6) =0.
Evaluate the value of t for the ball hit the ground.
[tex]\begin{gathered} -16t=0 \\ t=0 \end{gathered}[/tex]
OR
[tex]\begin{gathered} t-6=0 \\ t=6 \end{gathered}[/tex]
The ball height is 0 at t = 0 sec and at t = 6 sec.
The t = 0 sec represent the initial condition and t = 6 sec represents the time after which ball hit the ground again.
So answer is t = 6 seconds.
PART B.
Differentiate the equation and equate to 0 for maximum height.
[tex]\begin{gathered} \frac{d}{dt}(h)=\frac{d}{dt}(-16t^2+96t) \\ \frac{dh}{dt}=-32t+96 \end{gathered}[/tex]
For maximum height,
[tex]\begin{gathered} -32t+96=0 \\ t=-\frac{96}{-32} \\ t=3\text{ sec} \end{gathered}[/tex]
So part B answer is 3 sec
Answer:
Part A: 6 seconds
Part B: 3 seconds
