If an object is propelled upward from a height of 128 feet at an initial velocit
ID: 3149284 • Letter: I
Question
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per second, then its height after t seconds is given by the equation h(t) = -16t2 + 112t + 128.
a) After how many seconds does the object hit the ground? Round to the nearest second.
b) After how many seconds does the ball reach a height of 145 feet? Round your answers to the nearest hundredth of a second.
c) After how many seconds does the object reach its maximum height? Round to the nearest tenth of a second.
d) What is the maximum height that the object reaches? Round to the nearest foot.
Explanation / Answer
h(t) = -16t^2 + 112t + 128
a) when the object hits the ground h(t) = 0
-16t^2 + 112t + 128 = 0
-16 ( t^2 - 7t - 8 ) = 0
t = 8 seconds
object hits the ground after 8 seconds
b) 145 = -16t^2 + 112t + 128
-16t^2 + 112t - 17 = 0
t = [ -112 +- sqrt ( 112^2 - 1088) ] / - 32
t = 0.16 , 6.85 seconds
c) ball reaches maximum height after
t = - 112 / 2 (-16)
t = 3.5 seconds
d) maximum height = -16(3.5)^2 + 112(3.5) + 128
= 324 feet