A ball is dropped from the top of a building. The height, y, of the ball above the ground (in feet) is given as a function of time, t, (in seconds) by
y = 1210 -16^2

(a) Find the velocity of the ball at time t.
(b) When does the ball hit the ground, and how fast is it going? Give your answer in feet per second and miles per hour (1 ft/sec = 15/22 mph). Round your answers to two decimal places.

t = ?
v(t) = ? feet per second
v(t) = ? feet per hour

Question

A ball is dropped from the top of a building. The height, y, of the ball above the ground (in feet) is given as a function of time, t, (in seconds) by 
y = 1210 -16^2

(a) Find the velocity of the ball at time t. 
(b) When does the ball hit the ground, and how fast is it going? Give your answer in feet per second and miles per hour (1 ft/sec = 15/22 mph). Round your answers to two decimal places. 

t = ?
v(t) = ? feet per second
v(t) = ? feet per hour

anonymous · Answer

Well, since I'm learning about derivatives, I assume I need to use that to get my answer. 

I tried:
y = 1210 - 16t^2
= 1210 -32t
-1210 = -32t
t = - 32.81 

I know the sign has to be negative because the height is decreasing but... that's all I got.

ganeshie8 · Answer

is that for part a ?

anonymous · Answer

Yeah, I was trying to do part a. 
I'm guessing I need to apply a for b, so I didn't try that yet.

ganeshie8 · Answer

part a, is just asking u to get velocity as a function of t; thats all.

ganeshie8 · Answer

it should be below :-

y = 1210 - 16t^2
v = y' = -32t

anonymous · Answer

Alright, got it! 
Now I need to figure out how to apply this to (b), which I'm getting wrong as well.

ganeshie8 · Answer

to apply it to part b,
first u need to find time, t, at which ball hits the ground

ganeshie8 · Answer

any idea how to find that ? (Algebra)

ganeshie8 · Answer

|dw:1382418247717:dw|