Asked by jim
A photographer in a helicopter ascending vertically at a constant rate of 1.2m/s accidentally drops a camera out of the window when the helicopter is 46m above the ground. How long will it take the camera to reach the ground
Answers
Answered by
Damon
Hi = 46
Vi = 1.2
a = g = -9.81 m/s^2
h = Hi + Vi t - 4.9 t^2
0 = 46 + 1.2 t - 4.9 t^2
t^2 - .245 t - 9.39 = 0
t = [ .245 +/- sqrt ( .06 + 37.6) ]/2
t = [ .245 + 6.13 ] /2 forget negative time
t = 3.2 seconds
Vi = 1.2
a = g = -9.81 m/s^2
h = Hi + Vi t - 4.9 t^2
0 = 46 + 1.2 t - 4.9 t^2
t^2 - .245 t - 9.39 = 0
t = [ .245 +/- sqrt ( .06 + 37.6) ]/2
t = [ .245 + 6.13 ] /2 forget negative time
t = 3.2 seconds
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