A rocket moves upward, starting from rest with an acceleration of +29.4 m/s^2 for 3.98 sec. It runs out of fuel at the end of the 3.98 sec but does not stop. How high does it rise above the ground?
Answer is 931 m but idk how to get that. Please explain! 10 points is up for grabs!
Answer is 931 m but idk how to get that. Please explain! 10 points is up for grabs!
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Work the problem in two parts.
In part 1 the acceleration is constant so use your kinematic eqs to find the distance traveled "y1" and the velocity "v1" after 3.98 s;
y1 = (1/2)at^2
v1 = at
In part 2 the rocket is basically a projectile under the sole influence of gravity, with an initial velocity vo = v1 , found above. Again use your kinematic eqs to find how high it goes "y2" as a projectile;
v^2 = vo^2 - 2gy2
You find the height "y2" by solving for v=0 , when the rocket stops going up and is ready to fall back;
0 = v1^2 - 2gy2
y2 = v1^2/2g
The total height traveled by the rocket is then;
y = y1 + y2
In part 1 the acceleration is constant so use your kinematic eqs to find the distance traveled "y1" and the velocity "v1" after 3.98 s;
y1 = (1/2)at^2
v1 = at
In part 2 the rocket is basically a projectile under the sole influence of gravity, with an initial velocity vo = v1 , found above. Again use your kinematic eqs to find how high it goes "y2" as a projectile;
v^2 = vo^2 - 2gy2
You find the height "y2" by solving for v=0 , when the rocket stops going up and is ready to fall back;
0 = v1^2 - 2gy2
y2 = v1^2/2g
The total height traveled by the rocket is then;
y = y1 + y2