A 7400 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s^2 and feels no appreciable air resistance. When it has reached a height of 585 mm, its engines suddenly fail so that the only force acting on it is now gravity. Required:
a. What is the maximum height this rocket will reach above the launch pad?
b. How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes?
c. Sketch ayât,,vyâât, and y - t graphs of the rocketâs motion from the instant of blast-off to the instant just before it strikes the launch pad.

Explanation:

Let us first calculate the velocity of rocket on reaching the height of 585 m

v² = u² + 2 a s

= 0 + 2 x 2.35 x 585

v = 52.43 m /s

For onward journey of rocket , initial velocity u = 52.43 m /s , acceleration = - g = - 9.8 m /s² final velocity at top point v = 0

v² = u² - 2gh

0 = 52.43² - 2 x 9.8 x h

h = 140.25 m

Total height reached

= 585 + 140.25

= 725.25 m

b )

Time to reach the top position be t₁

v = u - gt₁

0 = 52.43 - 9.8 t₁

t₁ = 5.35 s

Time to reach the ground after reaching highest point be t₂

h = ut + 1/2 g t²

u = 0 as at top point , velocity = 0

725.25 = 0 + .5 x 9.8 t₂²

t₂² = 148

t₂ = 12.16 s

Total time = t₁ + t₂

= 5.35 + 12.16

= 17.51 s .

this could be newton#'s law of cooling. the cooling curve is exponential