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.
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.
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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
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
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h1 = ½a*t² = 250 m
v1 = a*t = 500 m/s
Hmax = h1 + v1²/(2*g) = 250 + 500²/(2*9.8) = 130005 m
v1 = a*t = 500 m/s
Hmax = h1 + v1²/(2*g) = 250 + 500²/(2*9.8) = 130005 m