A rocket is launched straight up. It contains two stages (Stage 1 and Stage 2) o

Question

A rocket is launched straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 10.0 and 5.0 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides a net upward acceleration of 18 m/s^2. In Stage 2, the net upward acceleration is 10 m/s^2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return back to the surface. maximum altitude 4621.23 m time required 76.88 s A rocket with two stages of rocket fuel is launched straight up into the air from rest. Stage 1 lasts 10.0 s and provides a net upward acceleration of 18 m/s^2. Stage 2 lasts 5.0 s and provides a net upward acceleration of 10 m/s^2. After Stage 2 finishes, the rocket continues to travel upward under the influence of gravity alone until it reaches its maximum height and then falls back towards Earth. We can split the rocket's flight into four parts: (1) Stage 1, (2) Stage 2, (3) between the end of Stage 2 and reaching the maximum height, (4) falling from the maximum height back to the Earth's surface. The acceleration of the rocket is constant over each of these legs, so we can use the constant acceleration equations to determine the total distance covered in each leg and the duration of each leg. The initial speed for legs #1 and #4 is zero, but we will need to calculate the initial speeds for legs #2 and #3. The maximum altitude is equal to the distance covered in legs #1 - #3: the time required for the rocket to return to Earth is equal to the total duration of its flight, which is the sum of the durations of legs #1 - #4.

Explanation / Answer

At stage 1, the final velocity of the rocket -

v1 = u + a1*t1 = 0 + 18*10 = 180 m/s.

at the stage2, the final velocity of the rocket -

v2 = v1 + a2*t2 = 180 + 10*5 = 230 m/s.

So, time taken by the rocket to become its velocity zero -

0 = 230 - 9.81*t3

=> t3 = 230 / 9.81 = 23.45 s.

So, the total time taken by the rocket in the upward journey -

T1 = t1 + t2 + t3 = 10+5+23.45 = 38.45 s.

Now, you have determined the maximum height attained by the rocket is 4621.23 m

So, we determine the time taken by the rocket to cover this downward distance.

4621.23 = 0 + (1/2)*9.81*t^2

=> t^2 = 942.15

=> t = 30.69 s.

Therefore, the time required by the rocket to return back to the earth's surface = T + t = 38.45 + 30.69 = 69.14 s.

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