A child launches a rocket, This rocket has 3 stages. In stage one the rocket lau

Question

A child launches a rocket, This rocket has 3 stages. In stage one the rocket launches from rest with an acceleration of 25 m/s^2 upward for 12 seconds. AI the start of stage 2 the rockets engines will turn off. In stage 3 the rocket falls back to the ground. A) What height does the rocket reach at the end stage 1? B) What velocity does the rocket reach at the end of stage 1? C) How much time must pass in stage 2 before the rocket stops moving upward? D) What is the maximum height the rocket will reach?

Explanation / Answer

Stage 1 :

initial velocity = Vi = 0 m/s

acceleration = a = 25 m/s2

time = t = 12 sec

d1 = distance travelled in stage 1

A)

using the equation ::

d1 = Vi t + (0.5) at2

d1 = (0)(12) + (0.5) (25) (12)2

d1 = 1800 m

B)

using the equation

Vf = vi + at

Vf = 0 + 12 x 25

Vf = 300 m/s

3)

for stage 2:

initial velocity , Vi = 300 m/s

final velocity , Vf = 0 m/s

acceleration , a = -9.8 m/s2

using the equation

Vf = Vi + at

0 = 300 + (-9.8)t

t = 30.6 sec

d)

height reached in stage 2 can be given as ::

Vf2 = Vi2 + 2 a h

02 = 3002 + 2 (-9.8) h

h = 4591.84 m

height reached in stage 1 = d = 1800

so total height = 4591.84 m + 1800 = 6391.84 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.