Problem 2.48 A rocket rises vertically, from rest, with an acceleration of 3.2 m

Question

Problem 2.48

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 850 m . After this point, its acceleration is that of gravity, downward.

Part A What is the velocity of the rocket when it runs out of fuel? Express your answer to two significant figures and include the appropriate units. v850m = 74 ms

Part B

How long does it take to reach this point?

Part C

What maximum altitude does the rocket reach?

Part D

How much time (total) does it take to reach maximum altitude?

Part E

With what velocity does it strike the Earth?

Part F

How long (total) is it in the air?

Explanation / Answer

(A) Applyng vf^2 - vi^2 = 2 a d

v1^2 - 0^2 = 2 x 3.2 x 850

v1 = 74 m/s

(B) Applying vf = vi + a t

74 = 0 + 3.2 t

t = 23 sec

(C) after that, a = - 9.8 m/s^2

at maximum altitue, v = 0

Applying vf^2 - vi^2 = 2 a d

0^2 - 74^2 = 2 (-9.8) h

h = 278 m

maximum altitude = 850 + 278 = 1128 m

(d) 0 = 74 - 9.8 t

t = 7.6 sec

total time = 7.6 + 23 = 30.6 sec

(e) v^2 - 0^2 = 2(-9.8)(-1128)

v = 149 m/s downwards.

(f) 149 = 0 + 9.8t

t = 15.2 sec

total time of flight = 30.6 + 15.2 = 45.8 sec

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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