Step-by-step explanation:
Given that,
The quadratic equation that models the ball's height above the ground h, t seconds after it was thrown is given by :
[tex]h=-16t^2+108t+28[/tex] ...(1)
(a) For maximum height, put dh/dt = 0
So,
[tex]\dfrac{d(h)}{dt}=\dfrac{d}{dt}(-16t^2+108t+28)=0\\\\-32t+108=0\\\\t=\dfrac{108}{32}=3.37\ s[/tex]
Put the value of t in equation (1).
So,
[tex]h=-16(3.37)^2+108(3.37)+28\\\\h=210.24\ ft[/tex]
(b) When the ball reaches the ground,
h = 0
So,
[tex]-16t^2+108t+28=0\\\\t=7\ s[/tex]
So, the ball hits the ground in 7 seconds.
