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A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0 above the horizontal. The rocket is fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

$h=33.22 \mathrm{m}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

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so the problem states that a rocket is launched at 75 meters per second, an angle of 60 degrees, and we're trying to find how high over this 11 meter high barrier the rocket clears on the barriers located 27 meters away from the rocket. So the first thing we really should do is figure out how long it takes the rocket to reach the barrier. To do this, we can separate the velocity into two components. So in this case we should really focus on the horizontal component of the Lost E. And we can find this because we know that the co sign of 60 degrees is going to be equal to the horizontal component of lossy time are divided by 75 meters per second, which means the of X is equal to 75 times co sign of 60 degrees. Now that we know the horizontal component of the velocity, we can plug it into our equation, which states that the change in displacement is equal to the initial velocity times time. The initial velocity in this case is just the horizontal velocity and so we can plug in 27 meters as well as 75 times co sign of 60 degrees and sulfur t And when we do this, we find that T is equal 2.72 seconds. Now that we know the time it takes the rocket to reach the barrier, we can figure out the height of the rocket at this certain time. To do this, we can use our kinetics formula which states that the change in displacement is equal to the initial velocity. Times that time plus one have times the acceleration times, the Times Square. In this case, we need to focus on the vertical component of the lost E. So be supply, which is we confined because we no sign of 60 degrees is equal to visa. Why over 75 meters per second because of so Kitona opposite overhype on news. And we find that visa Y is equal to 75 times the sign of 60 degrees. Now that we know this week unplug everything we know into the formula, we're trying to find Delta X. Of course we know what V not is 75 times sign of 60 degrees. We know what the time is. 0.72 seconds. We know that the acceleration will be negative 9.8 meters per second squared due to gravity and once again we know what the time is. We square this and then when we saw for the change in displacement, we get the displacement is 44.22 meters. But we're trying to find how much it clears this wall, so we need us attract off 11 meters and that gives us a clearance of 33.22 meters and that's the answer.

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