Is there anyone good @ Calculus??????

Question

I dont know how to solve this problem:

With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approximately 550 feet).
acceleration due to gravity: a(t) = -32 feet per second per second (neglect air resistance)

Answer 1

550= 1/2 *g * t^2 (coming down equation)
solving for t = 5.844 seconds

550= v * t - 1/2 g*t^2 (going up equation)

v= 550+1/2 32.2 * 5.84^2/ 5.84= 188 ft. per sec.

Answer 2

Its acceleration at time t is -32.
So integrate to get its velocity as -32t + c.
Then integrate again to get distance: -16t^2 + ct + d.
Now, at time 0 distance is 0, so d = 0.
It reaches maximum height when velocity is 0, ie -32t + c = 0, so 32t = c and t = c/32.
So its maximum height is -16(c/32)^2 + c(c/32)
= c^2 / 64.
So we want c^2 / 64 = 550, so c = sqrt(35200) = 187.6, approximately.

This is the initial velocity as substituting t = 0 into -32t + c gives c.

So 187.6 feet/s.

Answer 3

s=intial speed

speed at time t is (s-32t)
will each speed 0 (and max height) at t=s/32
position at time t: st-16t^2

plug in t=s/32 and set to 550:
s^2/32-s^2/64=550
s=sqrt(550*64) = ...

Answer 4

this is a physics problem...

vf^2 = vo^2 + as
0 = (v)^2 + (-32)(55)
1760 = v^2
v =42.0 m/s

Answer 5

CAN BE INFINITE.....

You can never reach it if you throw it upwards from Toronto....