Question

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.10 m/s2 . When it is 235 m 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). How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?What total distance did the canister travel between its release and its crash onto the launch pad? This includes the distance the canister travels while going up plus the distance it travels while falling back to the ground.

Answer 1

acceleration is 3.1 m/sec²

the speed is 235 m/s is V_f

using

V_f² - V_i² = 2 a s

V_f² - 0² = 2 X 3.1 X 235

V_f = 38.17 m/s

the maximum height is dh = V_f² / 2 g

dh = 38.17² / 2 X 9.8

dh = 74.33 m _________________ i

time to reach height is t₁ = V_f / g

= 38.17 / 9.8

t₁ = 3.895 sec

the max height is H = 235 + 74.33 = 309.33 m _____________ ii

for t₂ :

t₂ = ( 2 H / g )^1/2

t₂ = ( 2 X 309.33 / 9.8 )^1/2

t₂ = 7.945 sec

total time is t = t₁ + t₂

t = 3.895 + 7.945

t = 11.84 sec

the distance travelled by rocket is S_rocket = V_f t + 1.2 a t²

S_rocket = 38.17 X 11.84 + 0.5 X 3.1 X 11.84 X 11.84

S_rocket = 669.22 m

so the rocket height from launch pad

H ' = 669.22 +235

H ' = 904.22 m

then the total distance is D

so from i and ii

D = dh + H

= 74.33 + 309.33

D = 383.66 m