# In a chase, a car runs horizontally off a cliff at a speed of 31 m/s. How far away from the cliff will the car land if the car's vertical velocity is 42 m/s on impact with the ground?

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### 1 Answer

After the car leaves the cliff into air, it will behave like a projectile thrown horizontally off a height.

Initially, the vertical component of velocity is zero (the car leaves the cliff horizontally); the initial velocity is solely its horizontal component.

During its flight in air, the horizontal component of its velocity will remain unchanged while the vertical component would undergo acceleration at the rate of g.

Let the time required to travel to the point of its landing be t.

Then, for the vertical component of its velocity,

`42=0+9.81*t`

`rArr t=42/9.81=4.281346 s`

Vertical distance (height) covered by this time,

`h=0*t+1/2*g*t^2`

`=1/2*(9.81 m/s^2)*(4.281346 s)^2`

`=89.91 m`

Horizontal distance covered by this time,

=constant velocity*time

`=(31 m/s)*(4.281346 s)`

`=132.72 m`

So, the car will land at a horizontal distance of 132.72 m from the base of the cliff, and at a point which is 89.91 m lower.

Aerial distance of the point of landing of the car from the cliff (i.e. from the point of its leaving the cliff) is

`=sqrt(132.72^2+89.91^2)`

`=160.31 m.`

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