PHYSICS!!!

anonymous

2008-09-12 8:56 pm

A rocket moves straight upward, starting from rest with an acceleration of 9.4 m/s^2 (with gravity). It runs out of fuel at the end of 4.0 seconds and continues to coast upward before reaching a maximum height and falling back to earth.

a.) find maximum height about the ground.
b.) Find velocity of the rocket the moment before it hits the ground.

回答 (1)

Uncle Michael

2008-09-13 12:31 am

✔ 最佳答案

a)
Taking all upward quantities as positive.

Consider the journey of the first 4 seconds:
u = 0 m/s
a = 9.4 m/s2
t = 4 s

v = u + at
v = (0) + (9.4)(4)
Final velocity = 37.6 m/s

s = ut + (1/2)at2
s = (0)(4) + (1/2)(9.4)(4)2
s = 75.2 m

Consider the journey just after 4.0 s and just before the maximum height.
u = 37.6 m/s
a = -10 m/s2
v = 0 m/s

v2 = u2 + 2as
(0)2 = (37.6)2 + 2(-10)s
s = 70.7 m

The maximum height
= 75.2 + 70.7
= 145.9 m

=====
b)
Consider the journey just after 4.0 s and just before it hits the ground.
u = 37.6 m/s
a = -10 m/s2
s = -75.2 m

v2 = u2 + 2as
v2 = (37.6)2 + 2(-10)(-75.2)
v2 = 2917.76
v = -54 m/s

The velocity of the rocket just before it hits the ground
= 54 m/s (downwards)
=

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