Homework 7 section 5.1 (Homework)

Question

Evaluate the function to answer the question.

The height h, in feet, of a ball that is released 5 ft above the ground with an initial upward velocity of 90 ft/s is a function of the time t, in seconds, the ball is in the air and is given by the following.

h(t) = −16t² + 90t + 5, 0 ≤ t ≤ 5.68

a: Find the height of the ball (in feet) above the ground 2 s after it is released.

_____ ft

B: Find the height of the ball (in feet) above the ground 4 s after it is released.

_____ ft

Answer 1

h(2) = -16t^2 +90t +5

= -16(2^2) +90(2) + 5 = -64 + 180+ 5 = 121 feet high

f(4) = -16(4^2) + 90(4) + 5 = -256 + 365 = 109 feet high

max height is when -32t +90 = 0

t= 90/32 = 45/16 = about 2.8

max h = -16(2.8)^2 + 90(2,8) + 5 = -126.6 + 253.1 +5 = 131.5 feet

at 2 seconds the ball is going up, at 4 seconds it's going down.

at about 2.8 seconds, close to 3 or the middle of 2 and 4, it's at max height, so h(2) and h(4) are roughly equal. 121 and 109.