Aball 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

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

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

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

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 :

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

d

step-by-step explanation:

in a an geometric explicit equation, the first number is the starting rate. in this case 10.

i need the table so i can see

step-by-step explanation: