Question

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.45m/s2 . When it is 230m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).

1- How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?
2- What total distance did the canister travel between its release and its crash onto the launch pad?

Answer 1

Step-by-step explanation:

Time elapsed to reach the height of 230 m be t

s = ut + 1/2 at²

230 = .5 x 3.45 t²

t = 11.55 s

velocity at height of 230 m

v = u + at

= 0 + 3.45 x 11.55 = 39.84 m/s

This velocity will be attained by canister .

time to reach zero velocity at the top position t

v = u - gt

0 = 39.84 - 9.8 t

t = 4.06 s

height travelled by canister during this 4.06 s

v² = u² - 2gH

0 = 39.84² - 2 X 9.8 H

H = 80.98 M

Total height attained by canister = 80.98 + 230 = 310.98 m

Time of fall by canister t

s = 1/2 gt²

310.98 = .5 x 9.8 t²

t = 7.97 s

Total time taken by canister to reach the ground after its release from rocket

= 4.06 + 7.97 = 12.03 s

Distance travelled by rocket in 12.03 s

s = ut + 1/2 a t²

= 39.84 x 12.03 + .5 x 3.45 x 12.03²

= 479.27 + 249.64

= 728.91 m

height of rocket required = 230 + 728.91

= 958.91 m

2 )

Distance travelled by canister between its release and fall on the ground

= 80.98 + 80.98 + 230

= 391.96 m.