Question

A 7400 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.15 m/s2 and feels no appreciable air resistance. When it has reached a height of 550 m , its engines suddenly fail so that the only force acting on it is now gravity. a)What is the maximum height this rocket will reach above the launch pad? b)How much time after engine failure will elapse before the rocket comes crashing down to the launch pad? c)How fast will it be moving just before it crashes?

Answer 1

Step 1 -- to determine the velocity of the rocket when the engine failed. The working equation is

Vf^2 - Vo^2 = 2as

where

Vf = rocket velocity when the engine failed
Vo = initial velocity at launch pad = 0
a = 2.15 m/sec^2 (given)
s = distance travelled = 550 m (given)

Substituting appropriate values,

Vf^2 = 0 + 2(2.15)(550) = 2365

Vf = 48.63 m/sec.

Time taken :
v = u + at
48.63 = 2.15*t
t = 22.62 s

Step 2 -- determine the distance that the rocket will continue to travel until it reaches its maximum height

The working formula is the same as in Step 1 except for the following designations:

Vf = 0 (since rocket will stop when it reaches its maximum height)

Vo = 48.63 m/sec (as determined in Step 1)

a = g = acceleration due to gravity = 9.8 m/sec^2 (constant)

s = maximum height (after 550 meters) at which rocket will travel

Substituting appropriate values,

0 - (48.63)^2 = 2(-9.8)s

NOTE the negative sign attached to the acceleration due to gravity.
This simply implies that the rocket tends to slow down as it goes up after losing its engine power.

Solving for "s",

s = 48.63^2/(2 * 9.8)

s = 120.66 meters

Therefore, the maximum height that the rocket will reach above the launch pad is

a) 550 + 120.66 = 670.66 meters
b)
Time taken after engine failure
0 = 48.63 - 9.8*t
t = 48.63/9.8 = 4.96
Hence Total Time = 4.96 + 22.62 = 27.58 s

c) v = 48.63 m/s