Question

A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 48t + 6. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height?
A. 1.5 s; 54 ft
B. 1.5 s; 114 ft
C. 3 s; 6 ft
D. 1.5 s; 42 ft

Answer 1

Answer: D. 1.5 s; 42 ft

Step-by-step explanation:

Given : A  ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function

$h(t) = -16t^2 + 48t + 6$

To find the time taken to reach the maximum height by the ball , we differentiate the above function  with respect to 't', we get

$h' (t)=(2) -16t+ 48(1) + (0)=-32t+48$

Now put h' (x) =0 , we get

$-32t+48=0\\\\\Rightarrow\ t=\dfrac{48}{32}=\dfrac{3}{2}\\\\\Rightarrow\ t=1.5$

Hence, it will take 1.5 seconds to reach the maximum height by the ball.

The maximum height of the ball at t= 1.5 seconds will be :

$h(t) = -16(1.5)^2 + 48(1.5) + 6=-36+72+6=42 \text{ ft}$

Hence, the correct option is D. 1.5 s; 42 ft.