A two-stage rocket blasts off vertically from rest on a launch pad. During the f

Question

A two-stage rocket blasts off vertically from rest on a launch pad. During the first stage, which lasts for 16.0 s, the acceleration is a constant 2.00 m/s2 upward. At the end of the first stage the second stage engine fires, producing an upward acceleration of 3.00 m/s2 that lasts for 12.0 s. At the end of the second stage, the engines no longer fire and therefore cause no acceleration, so the rocket coasts to its maximum altitude. Over the time interval from blastoff at the launch pad to the instant that the rocket falls back to the launch pad. Ignore the effects of air resistance.

1. What is the maximum altitude of the rocket? (Express your answer to three significant figures.)

2. What is its average speed? (Express your answer to three significant figures.)

3. What is its average velocity? (Express your answer to three significant figures.)

P.S The answer for #1 is not 856m

Explanation / Answer

Given

Initial Velocity u = 0 m/s

First stage acceleration a1 = 2 m/s2

Second stage acceleration a2 = 3 m/s2

First stage duration t1 = 16.0 s

Second stage duration t2 = 12.0 s

Solution

A)

Altitude reached during First stage

H1  = ut1 + ½ a1t12

H1 = 0 x 0 + ½ x 2 x 162

H1 = 256 m

Speed at the end of the final stage

V1 = u + a1t1

V1 = 0 + 2 x 16

V1 = 32 m/s

Altitude reached during second stage

H2  = v1t2 + ½ a2t22

H2  = 32 x 12 + ½ x 3 x 122

H2  = 600 m

Speed at the end of the second stage

V2 = v1 + a2t2

V2  = 32 + 3 x 12

V2  = 104 m/s

Now at the end of the second stage, the rocket will be under the influence of gravity, the rocket will continue to move upward until it reaches a point where its kinetic energy is zero

Duration of this travel ( from the end of second stage to maximum altitude)

V3 = v2 - gt3

0 = 104 - 9.8 x t3

t3 = 10.6 s

Ditance ( from the end of second stage to maximum altitude)

H3  = v2t3 - ½ gt32

H3   = 104 x 10.6 - ½ x 9.8 x 10.62

H3  = 551.8 m m

(verification)

V32 = v22 - 2gH3

0 = 1042 - 2 x 9.8 x H3

H3 = 551.8 m

The maximum altitude reached

H = H1 + H2 + H3

H = 256 + 600 + 551.8

H = 1407.8 m

B)

After reaching the highest altitude the rocket will start to fall and reach the lauch pad on earth surface

Duration of this fall

H = v3t4 + ½gt42

1407.8 = 0 + ½ x 9.8 x t42

t4 = 17 s

Total duration of flight

T = t1 + t2 + t3 +t4

T = 16 + 12 + 10.6 + 17

T = 55.6 s

The total distance travelled by the rocketduring upward and downward course

S = H + H

S = 1407.8 + 1407.8

S = 2815.6 m

Average Speed

Vs = S/T

Vs = 2815.6 / 55.6

Vs = 50.6 m/s

C)

The rocket srats from the lauch pad and fall back to the lauch pad.

So thesplacement (shortest distance between starting and end points of journey) is zero

Since velocity = Displacement/ duration

Velocity is also zero

Vv = 0 m/s

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