Question

an object is thrown upward from the top of a tower. its height is above the ground in feet at t seconds after it is thrown is given by
f(t)= -16t^2 +128t+144
a. at what time does the object hit the ground
b. how fast is it going as it hits the ground
c. how high does it go
d. how tall is the tower
e. what is the objects velocity as it passes the top of the tower on its way down.

Please answer this, I need to know how to do it before my finals

Answer 1

So the f(t) you are given is a position function. Taking the derivative of that gives you velocity.

A. The object hits the ground when the position function equals 0. So you set f(t) = 0. After you factor it, you get (t-9)(t+1) which says that the position (or height) = 0 when t = 9 seconds and when t = -1 seconds. Just ignore the negative time value and your answer becomes t = 9 seconds.

B. So if you differentiate the position function f(t), you get f'(t) which is velocity. f'(t) = -32t +128. Plug in your time value for when it hits the ground, t = 9, then you get -160 ft/s. The negative indicates the downward direction.

C. When the ball reaches its max height, the velocity equals 0. So set your velocity function equal to 0. This being f'(t) = -32t+128 = 0. Solving this gives you t = 4. Plug that time into the position function f(t) and it gives you a max height of 400 ft.

D. At t=0, the ball is in your hands, and you are at the top of the tower. So to find out how tall the tower is, just plug t = 0 into f(t) and you should get 144 ft.

E. For this, just set f(t) = 144. f(t) = -16t^2 + 128t+144=144. Subtracting 144 over to the left side, you get f(t) = -16t^2 + 128t = 0. You can simplify this to t(-16t+128)=0. Then just solve for the t in parentheses (-16t+128=0) and you'll find t = 8 seconds.

Hope this helps and sorry for any typing errors I made.

Answer 2

I forgot to say on part E. once you get t = 8 seconds, plug that into f'(t) to get the velocity at 8 seconds.

Answer 3

Thank you so much