Question

A car accidentally rolls of cliff as it leaves the cliff it has a horizontal velocity of 15 m/s,it hits the ground 60m from the shore line calculate the height of the cliff

Answer 1

The height of the cliff is 78.4 m

Step-by-step explanation:

Horizontal Launch Motion

When an object is thrown horizontally with a speed v from a height h, it describes a curved path exclusively ruled by gravity until it hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

`v_x=v_o`

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

`v_y=g.t`

Where g is the acceleration of gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

`\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}`

If the horizontal distance and the speed are known, we can solve the equation above for h:

`\displaystyle h=(d^2g)/(2v^2)`

The car leaves a cliff at a horizontal speed of v=15 m/s and hits the ground d=60 m from the shoreline. Thus the height of the cliff is:

`\displaystyle h=(60^2*9.8)/(2*15^2)`

Calculating:

h = 78.4 m

The height of the cliff is 78.4 m