A helicopter is rising vertically at constant velocity of 10 m/s. When the helic

Question

A helicopter is rising vertically at constant velocity of 10 m/s. When the helicopter is at a height of 75m above the ground, a passenger accidentally drops her camera out of the window of the helicopter. Ignore air-resistance. Where a numerical value is necessary, take the acceleration due to gravity to be 10m/s2.

a. Calculate the velocity of the camera as it reaches the ground.

b. How many seconds pass before the camera hits the ground?

c. What is hte maximum height reached by the camera?

Explanation / Answer

A) Intial Velocity u =-10 Final Velocity = v g= 10 S= 75 V^2 = u^2 +2gs => v^2 = 100 + 2*10*75 => V = 40m/sec B) v= u +gt => 40 = -10+10*t =>t =5 sec C) For max height V= 0 let the height for the point of drop be h v^2 = u^2 +2gs => s= 5 total max height = 75+5= 80m

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