Landing Pad
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A model rocket is fired from a launch pad. The rocket’s engine burns for 6 seconds, providing an acceleration
...of 4m/s^2, and the rocket is aimed 12 degrees from the vertical. After the 6 second burn, the rocket eventually falls back to earth under the influence of gravity.
a) How long does the entire process take?
b) How far from the launch pad does the rocket land?
Hey, Country!
What you need to do is divide this problem into three parts:
Part I - the rocket is accelerating upward.
Part II - the engine stops burning, the rocket starts decelerating, but is still heading upward (toward its apex/apogee).
Part III - the rocket accelerates downward the ground.
Part I
the rocket's velocity when the engine stops burning: v = v(0) + a*t = 0 + 4*6 = 24 m/s. Its distance is x(0) + v(0)*t+(1/2)*a*t² = (1/2)*4*36 = 72 m.
We'll now take these two figures and break them down into horizontal and vertical components. It would probably help to draw a diagram w/a vertical line, and a second line 12°from that. Our final horizontal velocity (for Phase I)is 24*sin(12°) = 4.99 m/s, and our final vertical velocity is 24*cos(12°)=23.48m/s.
Our horizontal distance is 72*sin(12°)=14.97m, and our vertical distance is 72*cos(12°) = 70.43m.
Part II
The rocket is finished burning. It will continue to move horizontally at 4.99 m/s. It will, however, decelerate from 23.48 m/s upward to 0 m/s. We can use the same equation as before: 0 (our final velocity) = 23.48 + (-9.8)t. If you solve for t, you get t=2.4s. So, the rocket moved another 4.99 m/s * 2.4s=11.98m horizontally, and it moved another (1/2)*(9.8)*(2.4²)=28.22m upward.
The total height of the rocket is 70.43+28.22=98.65 m.
Part III
We've done most of the hard work up front: this last part should be a little easier. We need to calculate the time it takes the rocket to fall 98.65m. We can use that time to then calculate the horizontal distance traveled. With the apex as our reference we take the distance equation from earlier, x(0)=0, v(0)=0, so 98.65=(1/2)*(9.8)*t². 20.13=t², so t=4.49s. The horizontal distance is (4.49)*(4.99)=22.41m.
SUMMARY
We take all of our times and add them:
a.) 6 + 2.4 + 4.49 = 12.89s
We then take our horizontal distances and add those:
b.) 14.97 + 11.98 + 22.41 = 49.36 m.
I would use up a lot more space and time if I went into detail to explain why I used -9.8 in one place and +9.8 in others. If you're unsure, ask your teacher to explain it.
I hope this was helpful. Good luck.
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