SOLUTION: an arrow is shot vertically upward with a speed of 288ft/s, and 3.00 s later another is shot up at a speed of 240 ft/s. will they meet? If so, where?

Question 1132492: an arrow is shot vertically upward with a speed of 288ft/s, and 3.00 s later another is shot up at a speed of 240 ft/s. will they meet? If so, where?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
arrow is shot vertically upward with a speed of 288ft/s, and 3.00 s later another is shot up at a speed of 240 ft/s. will they meet? If so, where?
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Using 32 ft/sec/sec for gravity:
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will they meet?
The 1st arrows height h(t) = -16t^2 + 288t
At t = 3 seconds, h(3) = -16*9 + 288*3 = 720 feet
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Its speed = 288 - 3*32 = 192 ft/sec
From that point, its height = -16t^2 + 192t + 720
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The 2nd arrows height = -16t^2 + 240t
When the 2nd arrow is launched, the 1st arrows height = -16t^2 + 192t + 720
Find where the 2 heights are equal.
-16t^2 + 192t + 720 = -16t^2 + 240t
48t = 720
t = 15 seconds --- that's 15 seconds from the 2nd launch.
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For the 1st arrow: h = -16*18^2 + 288*18 = 0 feet AGL
For the 2nd arrow: h = -16*15^2 + 240*15 = 0
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They meet at impact, 18 seconds after the 1st launch

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