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A Parachutist with a camera, with descending at a speed of 12.5m/s, releases, the camera at an altitude of 64.3m. What is the magnitude of the velocity of the camera just before it hits the ground ? The acceleration of gravity is 9.8 m/s and air friction is negligible. How long does it take the camera to reach the ground

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Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :


h = ut+(gt^2)/(2)

Putting all given values, we get :


12.5t+(9.8t^2)/(2)=64.3\\\\4.9t^2+12.5t=64.3

t = 2.56 and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

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