A car accidently volls of f a cliff a is leaves the cliff has a horzontal velocity of 13m/s if hits the ground 6am from the share line calculate the heigh of the cliff

Complete question :

A car accident my rolls off a cliff as it leaves the cliff it has a horizontal velocity of 13 ms it hits the ground 60 m from the shoreline calculate the height of the cliff

104.59 m

Explanation:

Given that:

Horizontal Velocity = 13 m/s

Horizontal distance = 60 m

To obtain the height of the cliff :

We use the Relation :

S = ut + 0.5at²

t = time ; u = initial velocity ; S = height

To obtain t ;

Speed = distance / time

Horizontal distance / speed = time

Time = 60 / 13

Time = 4.62 seconds

The height of cliff :

u = 0

Hence ;

S = ut + 0.5at²

S = 0 + 0.5(9.8) * 4.62^2

S = 104.58756

Height of cliff = 104.59 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards