Question

A spaceship takes off at a const. acceleration from earth for 10 sec. and then continues upwards for another 10 secs. until it reaches the height of 5000 m with a velocity of zero at that point before it begins to fall to earth.

In the first 10 sec. what is the net acceleration the spaceship experienced.

What was the height when the engines turned off?

What was the acceleration did the spaceship’s engines apply for the first 10 sec. (Different answer then question 1)

If the spaceship had accelerated for the entire 20 secs., what would have been the height and velocity at 20 secs?

Answer 1

Let the net acceleration be 'a'. Let point A be the surface of earth where the engine starts accelerating, point B be when the engine shuts off, and point C be the maximum height of the spaceship.

h_B = at²/2 = a * 10² / 2 = 50a

v_B = at = 10a

Conservation of mechanical energy at B and C gives,

mv_B²/2 + mgh_B = mgh_C

=> v_B²/2 + gh_B = gh_C

=> (10a)²/2 + g(50a) = 5000g

=> 50a² + 50ag - 5000g = 0

=> a² + ga - 100g = 0

=> a = {-g + [g² + 400g]^1/2}/2

=> a = {-9.81 + [9.81² + (400 * 9.81)]^1/2}/2 = 26.8 m/s²

So, height when engines are turned off, h_B = 50 * 26.8 = 1340 m

Acceleration by the ship, a' = a + g = 26.8 + 9.81 = 36.6 m/s²

If the spaceship had accelerated for entire 20 s,

h = at²/2 = 26.8 * 20² / 2 = 5360 m

v = at = 26.8 * 20 = 536 m/s