Problema Solution
If an object is thrown straight up from the top of a 240 foot tall building at 88 feet per second, the height of the object after t seconds can be modeled by the equation h(t)= -16t^2+88t+240.
a) How long until the object hits the ground?
b) What is the maximum height of the object?
c) When will the object reach a height of 272 feet? Explain your anwer.
Answer provided by our tutors
h(t)= -16t^2+88t+240.
(a) At the ground, h(t)=0. So, 0=--16t^2+88t+240. Solving the quadratic equation: (t-15/2)(t+2)=0
=> t =15/2 =7.5 seconds.
(b) At maximum height, dh(t)/dt =0==> -32t+88=0 ==> t=88/32 = 2.75 seconds.
So, maximu height = h(2.75) = -16*(2.75)2 +88*2.75+240 = 361 feet
(c) AT a height 272 feet => 272 = -16t^2+88t+240.
==> -16t^2+88t-32 = 0 ==> 16t^2-88t+32 = 0.
Solving the quadratic equation ==> t = 5.11 seconds.