SOLUTION: A golf ball is launched upward with an initial speed of 30 m/sec by a golfer atop the Washington Monument, which is 160 m above the ground. How high above the ground will the ball

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Question 402443: A golf ball is launched upward with an initial speed of 30 m/sec by a golfer atop the Washington Monument, which is 160 m above the ground. How high above the ground will the ball be after 3 sec?
I'm not sure how to set this up. Should I use: distance = rate × time?

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Distance = rate*time only works for constant velocity problems. In this problem, we have the acceleration due to gravity.

The formula for determining position given a constant acceleration is X+=+%281%2F2%29at%5E2+%2B+v%5B0%5Dt+%2B+x%5B0%5D, where a = -9.81 m/s^2, v%5B0%5D is the original velocity, x%5B0%5D is the original position (this equation is derived using integral calculus). Here, t = 3 s, v%5B0%5D = 30 m/s, x%5B0%5D = 160 m. Therefore,

X+=+%281%2F2%29%28-9.81m%2Fs%5E2%29%283s%29%5E2+%2B+%2830m%2Fs%29%283s%29+%2B+160m = 205.9 meters