A photographer in a helicopter ascending vertically at a constant rate of 10.5 a

Question

A photographer in a helicopter ascending vertically at a constant rate of 10.5 accidentally drops a camera out the window when the helicopter is 56.0 above the ground.
A)How long will the camera take to reach the ground?
B)What will its speed be when it hits?

Explanation / Answer

Set initial values: Initial height of camera = 56 Initial velocity of camera = 0 Acceleration due to gravity = -9.8 ------------------------------------------------ Set up equations: a(t) = -9.8 v(t) = -9.8t h(t) = -4.9t^2 + 56 ------------------------------------------------ Solve for t when h(t) = 0 --> -4.9t^2 + 56=0 --> 4.9t^2 = 56 --> t^2=60/4.9=11.428 --> t=sqrt(11.428) --> t=3.3805325 Time of impact = 3.4 sec ----------------------------------------------------------- Substitute into velocity function and find v(3.4) Speed is always positive v=(-9.8)*(3.4)= -33.32 --> speed at impact = 33.3 m/s

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