What is the maximum height this rocket will reach above the launch pad?

Question

situation: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail so that the only force acting on it is now gravity.

Answer 1

Vf^2=Vi^2+2ad

Vi=0 so eliminate that term

Vf^2=2(2.25)(525)

Vf^2=(4.5)(525)

Vf^2=2362.5

Vf=sqrt. 2362.5

Vf=48.61 m/s

so thats the velocity at the point the engine shuts off

Vf^2=Vi^2+2ad

0=(2,362.5)+2(-9.8)d

0=(2,362.5)-19.6d

-2,362.5=-19.6d

-2,362.5/-19.6=d

120.54=d

so it is going to rise a total of 645.54 m

Answer 2

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 = 2.25 m/sec^2 (given)

s = distance travelled = 525 m (given)

Substituting appropriate values,

Vf^2 = 0 + 2(2.25)(525) = 2362.5

Vf = 48.60  m/sec.

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 = 48.60 m/sec (as determined in Step 1)

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

s = maximum height (after 525 meters) at which rocket will travel

Substituting appropriate values,

0 - (48.60)^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 = 48.60^2/(2 * 9.8)

s = 120.53  meters

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

525 + 120.53 = 645.53 meters