calculus!!!!!!!!!?

Question

A stone is thrown straight up from the edge of a roof, 675 feet above the ground, at a speed of 20 feet per second.
A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 3 seconds later? _____________feet
B. At what time does the stone hit the ground? ___________seconds
C. What is the velocity of the stone when it hits the ground? ____________feet/second

Answer 1

More physics than calculus...some helpful relationships are

y - yo = vo t + 1/2 a t²

v = vo + a t

2 a ( y - yo ) = v² - vo²

Let's put the origin of the coordinate system is at the bottom of the building and say that "up" is positive. Then

t = time
y = position at time t
yo = position at beginning of problem ( 675 ft )
g = local acceleration of gravity ( -32 ft / s² )
v = velocity at time t
vo = initial velocity ( 20 ft / s )

(A) Use the first equation; plug in the values and solve for y.
(B) Use the first equation and set y = 0; solve for t.
(C) Use that time in the second equation and solve for v, or use the third equation instead.

Answer 2

i'm not going to do your homework for you but i will tell you how to do it.
1) use v=u+at and integrate with respect to t to get an expression for s.
2 use the eqn you derived in 1 to solve it. s=-675ft
3) lots of ways to answer this. personally, i'd work out the height when the stone is at it's highest point, from that it's very easy to work out velocity using v=at

Answer 3

not my thing