Question

A model jet is fired up in the air from a 16-foot platform with an initial upward velocity of 52 feet per second . The height of the jet above the ground after t seconds is given by the equation h=-16t^2+52t+16 where h is the height of the jet in feet and t is the time in seconds since it is launched . What is the maximum height the jet reaches , to the nearest foot
A) 777 feet
B)58 feet
C)30 feet
D)16 feet

Answer 1

Answer:

B.    58 feet to nearest foot.

Step-by-step explanation:

h = -16t^2 + 52t + 16

Convert to vertex form:-

= -16 [(t^2 - 3.25t) + 16

= -16 [ (t - 1.625)^2  - (1.625)^2 ] + 16

= -16 (t - 1.625)^2 + 42.25 + 16

= -16 (t - 1.625)^2  + 58.25.

The maximum height occurs when t = 1.625 seconds and its equal to 58 feet.