SOLUTION: a photographer in helicopter ascending vertically at a constant rate of 12.5m/s accidentally drops a camera out the window when the helicopter is 60m above the ground,how long will

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Question 490849: a photographer in helicopter ascending vertically at a constant rate of 12.5m/s accidentally drops a camera out the window when the helicopter is 60m above the ground,how long will the camera take to reach the ground,with what speed will the camera hit the ground and how high will the helicopter be when the camera hits the ground
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Set initial values:
Initial height of camera = 60
Initial velocity of camera = 0
Acceleration due to gravity = -9.8
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Set up equations:
a(t) = -9.8
v(t) = -9.8t
h(t) = -4.9t^2 + 60
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Solve for t when h(t) = 0
--> -4.9t%5E2+%2B60+=+0
--> 4.9t%5E2+=+60
--> t%5E2+=+60%2F4.9+=+12.245
--> t+=+sqrt%2812.245%29
--> t+=+3.5
Time of impact = 3.5 sec
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Substitute into velocity function and find v(3.5)
--> v+=+%28-9.8%29%2A%283.5%29+=+-34.3
Speed is always positive
--> speed at impact = 34.3 m/s
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Helicopter is ascending at a rate of 12.5 m/s
time = 3.5
distance = 12.5*3.5 = 43.75
--> 60+43.75 = 103.75
Helicopter is 103.75 m high at impact