Simple kinematics help
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Simple kinematics help

[From: ] [author: ] [Date: 11-05-12] [Hit: ]
It should be simple kinematics, but I am totally lost.Would you please show your work?Thanks.-a= 5m/s/s for 10 s,initial velocity is zero.......
A rocket, starting from rest, is launched vertically upwards with an acceleration of 5 m/s^2. After 10 s
the engine is switched off. Calculate the maximum height above the launch point reached by the rocket
before it begins its downward descent.
It should be simple kinematics, but I am totally lost. Would you please show your work? Thanks.

-
a= 5m/s/s for 10 s, then a= -10m/s/s
initial velocity is zero. V(f)= V(o) + at
V(f) = 0 + 5 * 10
V(f) = 50 m/s after 10 sec.

The height of the rocket when the engine switches off can be found using
V(f)^2 = V(o)^2 + 2ay
50^2 = 0 + 2 * 5 * y
250 = y.

Then gravity takes over, so a = -10, and the final velocity becomes the initial velocity for the second portion. We're finding the maximum height, when the velocity is zero, so the new V(f) is 0

V(f)^2 = V(o)^2 + 2ay
0 = 50^2 + 2 * -10 * y
-2500 = 2 * -10 * y
-1250 = -10 * y
125 = y

The rocket's maximum height is 250 meters of rocket flight + 125 meters of free flight = 375 meters

-
h1 = ½a*t² = 250 m
v1 = a*t = 500 m/s

Hmax = h1 + v1²/(2*g) = 250 + 500²/(2*9.8) = 130005 m
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