Question

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.15 m/s2 . When it is 260 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). A) How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? B)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

a) v_f = sqrt[2 a s] = sqrt [2 * 3.15 * 260] = 40.47 m/s

delta h = v_f² / 2 g = 40.47²/2*9.8 = 83.56 m

t₁ = v_f / g = 40.47 / 9.8 = 4.13 s

H = 260 + 83.56 = 343.56 m

t₂ = sqrt [2 H / g] = sqrt [2 * 343.56 / 9.8] = 8.37 s

t = 4.13 + 8.37 = 12.5 s

D = v_f t + 1/2 a t² = (40.47 * 12.5) + (1/2 * 3.15 * 12.5²) = 752 m

H' = 752 + 260

= 1012 m

b) total distance = 343.56 + 83.56

= 427 m

Answer 2

A) To find how high the rocket is when the canister hits the launch pad, we can use the equations of motion. The key point to note is that the acceleration of the rocket remains constant at 3.15 m/s², even after the canister is discarded.

Let's define:

• Initial height of the rocket above the launch pad, h_initial = 0 m (since it starts from rest on the launch pad).

• Height of the rocket when the canister is discarded, h_canister_discard = 260 m.

• Acceleration of the rocket, a = 3.15 m/s².

We need to find the height of the rocket (h_final) when the canister hits the launch pad. To do this, we can use the following equation of motion:

h_final = h_initial + v_initial*t + (1/2)at²

where:

• v_initial is the initial velocity of the rocket when the canister is discarded.

• t is the time it takes for the canister to hit the launch pad.

Since the rocket starts from rest, v_initial = 0.

Now, we need to find the time it takes for the canister to hit the launch pad. To do this, we can use the following equation of motion:

h_canister_discard = (1/2)at²

Solving for t:

t = sqrt(2*h_canister_discard / a)

Substitute the values:

t = sqrt(2*260 m / 3.15 m/s²) ≈ 8.15 seconds

Now, we can find the final height of the rocket:

h_final = h_initial + v_initial*t + (1/2)at² h_final = 0 + 0 + (1/2)*3.15 m/s² * (8.15 s)² h_final ≈ 0 + 0 + 103.83 m h_final ≈ 103.83 meters

Therefore, the rocket is approximately 103.83 meters high when the canister hits the launch pad.

B) To find the total distance the canister traveled between its release and its crash onto the launch pad, we need to consider two parts: the distance traveled while going up and the distance traveled while falling back down.

1. Distance traveled while going up: The distance traveled while going up is equal to the initial height of the rocket when the canister is discarded, which is 260 meters.

2. Distance traveled while falling back down: The time taken to reach the launch pad can be found using the value of t calculated above (t ≈ 8.15 seconds). During this time, the canister is in free fall due to gravity. The distance traveled while falling back down can be found using the equation of motion:

distance_falling = (1/2)gt²

where:

• g is the acceleration due to gravity, which is approximately 9.81 m/s².

Substitute the values:

distance_falling = (1/2)*9.81 m/s² * (8.15 s)² distance_falling ≈ 320.87 meters

Now, we can find the total distance traveled by adding the distance traveled while going up and the distance traveled while falling back down:

total_distance = distance_going_up + distance_falling total_distance ≈ 260 meters + 320.87 meters total_distance ≈ 580.87 meters

Therefore, the total distance the canister traveled between its release and its crash onto the launch pad is approximately 580.87 meters.