Question

A rocket is shot upward from a 20 foot tall tower at 564 feet/sec. What is it's maximum height, when does it hit the ground and what is it's velocity at 10 seconds?

Answer 1

maxm. height is when final velocity = 0
v^2 = u^2 + 2 a s
0 = 564^2 - 2 * 32 s
s = 564^2 / 64
= 4970. 25 ft above the tower

time it hits ground
use the equation
s = ut + 0.5at^2
s = displacement = - 20 ft, a = -g = -32, t = ?

-20 = 564t - 0.5*-32 t^2
16t^2 + 564t + 20 = 0
solve for t to give time in seconds it hits the ground

Answer 2

error - equation is 16t^2 - 564t - 20 = 0
t = 35.3 secs

Answer 3

velocity at 10 secs:

t=10, u = 564, v = ?, a = -32

v = u + a t
v = 564 - 32*10
v = 564 - 320 = 240 ft / sec = velocity after 10 seconds