Answer:
the maximum height of the ball is 42.25 m
Step-by-step explanation:
Given the height function as;
h(t) = -16t² + 12t + 40
At maximum height, the final velocity of the ball will be zero.
The final velocity is calculated as follows;
[tex]v = \frac{dh(t)}{dt} = -32 t+ 12\\\\at \ maximum \ height\ v = 0\\\\Thus, -32 t+ 12 = 0\\\\32t = 12\\\\t = \frac{12}{32} \\\\t = 0.375 \ s[/tex]
At maximum height, the time of motion of the ball is 0.375 s.
The maximum height is calculated as follows;
h(t) = -16t² + 12t + 40
h(0.375) = -16(0.375)² + 12(0.375) + 40
h(0.375) = -2.25 + 4.5 + 40
h(0.375) = 42.25 m
Therefore, the maximum height of the ball is 42.25 m
