A rock fell of the edge of a cliff and encountered a displacement of 50m. The final vertical velocity of the rock was 10m/s. Find the time the rock was in the air, the horizontal velocity when the rock left the cliff and the vertical displacement the rock had.

A.) 3.19s

B.) 0

C.) 55.1

Explanation:

Given that a rock fell of the edge of a cliff and encountered a displacement of 50m. The final vertical velocity of the rock was 10m/s.

To Find the

A.) time the rock was in the air,

Given that

h = 50m,

initial velocity U = 0,

acceleration due to gravity g =9.8m/s^2

Using second equation of motion:

h = Ut + 1/2gt^2

Substitute all the parameters into the equation

50 = 1/2 × 9.8 × t^2

100 = 9.8t^2

t^2 = 100/9.8

t^2 = 10.2

t = 3.19s

B.) the horizontal velocity when the rock left the cliff

The horizontal velocity is equal to zero since horizontal acceleration is zero.

C.) and the vertical displacement the rock had.

Using the third equation of motion.

V^2 = U^2 + 2gH

Where U = 0

V = 10m/s

Substitute all the parameters into the formula

10^2 = 0 + 2 × 9.8 × H

100 = 19.6H

H = 100/19.6

H = 5.1

The vertical displacement = 50 + 5.1

The vertical displacement = 55.1 m

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answer

1,500

kg • m/s.