Question

The height of an object thrown upward with an initial velocity of 96 feet per second is given by the formula h = −16t2 + 96t, where t is the time in seconds.
a)  At which time(s) will it take the object be at a height of 128 feet?     

b)  How long will it take the object to return to the point of departure?     

c)  How long will it take the object to reach its maximum height?

d) What is the Maximum Height?  

Answer 1

Given that the height of the object is h(t)=-16t^2+96t
a] At which time(s) will it take the object be at a height of 128 feet? 
To solve this we proceed as follows:
we need to solve for t at h(t)=128
thus
-16t^2+96t=128
rewriting the above we get:
-16t^2+96t-128=0
simplifying gives us:
t^2-6t+8=0
factoring the quadratic we get:
t^2-4t-2t+8=0
t(t-4)-2(t-4)=0
(t-2)(t-4)=0
thus
t=2 and t=4
hence the object will be at height 128 at t=2 sec and t=4 sec

b]How long will it take the object to return to the point of departure?
Here shall proceed as follows:
At the point of the departure, h(t)=0
thus plugging this in the equation we get:
-16t^2+96t=0
simplifying this we get:
-16t^2=-96t
dividing through by -16 we get
t^2=6t
hence
t=6 sec
The time taken for the object to return to the departure is 6 sec

c] How long will it take the object to reach its maximum height?
Time taken to reach the maximum height will be calculated as follows:
h(t)=-16t^2+96t
at maximum height h'(t)=0
but
h'(t)=-32t+96
equating this to zero we get:
-32t=-96
thus
t=3 sec
thus time taken to reach the height will be
t=3 sec

d]What is the Maximum Height? 
To get the maximum height we plug t=3 sec and simplify.
h(t)=-16t^2+96t
h(3)=-16(3)^2+96(3)
h(3)=-48+288
h(3)=240 ft