Answer:
After 1.5 second of throwing the ball will reach a maximum height of 44 ft.
Step-by-step explanation:
The height in feet of a ball after t seconds of throwing is given by the function
h = - 16t² + 48t + 8 .......... (1)
Now, condition for maximum height is
[tex]\frac{dh}{dt} = 0 = - 32t + 48[/tex] {Differentiating equation (1) with respect to t}
⇒ [tex]t = \frac{48}{32} = 1.5[/tex] seconds.
Now, from equation (1) we get
h(max) = - 16(1.5)² + 48(1.5) + 8 = 44 ft.
Therefore, after 1.5 seconds of throwing the ball will reach a maximum height of 44 ft. (Answer)
