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

Answer:

The height of the cliff is 78.4 m

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:

$\displaystyle h=\frac{d^2g}{2v^2}$

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:

$\displaystyle h=\frac{60^2*9.8}{2*15^2}$

Calculating:

h = 78.4 m

The height of the cliff is 78.4 m