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| Math Central | Quandaries & Queries |
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Question from leria, a student: equation for motion of moon is h=2.67t^2 + vt + s Suppose you are on an outpost on the moon and Jan is back on Earth. Both people toss a ball from a height of 96 feet with an initial velocity of 16 feet per second. How much longer will your ball stay in motion on the moon than Jan's on the earth? Math Central, I'm having a difficult time understanding which numbers to plug into the formula. |
Hi Leria.
The equation you wrote for the earth should be h = -16t2 + vt + s and the equation for the moon should be h = -2.67t2 + vt + s.
"s" is the starting height. "v" is the initial velocity (positive v means upwards). "h" is the height of the projectile (ball).
The ball starts from 96 feet and lands at 0 feet (you are throwing the balls off of a cliff, perhaps). You are throwing the ball straight up.
Thus your two equations become:
0 = -2.67t2 + 16t + 96 and 0 = -16t2 + 16t + 96.
These are two different times, so really two different "t" variables. So let e be the time it takes on the earth and m be the time it takes on the moon.
The question asks how much longer the ball stays in motion (I presume it is a glass ball that breaks when it hits the surface, so there is no bouncing!) on the moon than on the earth. That's m - e.
So you have
0 = -2.67m2 + 16m + 96 and 0 = -16e2 + 16e + 96.
and you want m - e.
Use any method, such as factoring or the quadratic formula or completion of squares to determine each of e and m, then subtract to answer the question.
Cheers,
Stephen La Rocque.
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