Question

A car rolls off a horizontal cliff 125 meters high, and lands 100 meters from the base of the cliff. How fast was the car moving when it rolled off the cliff?

Answer 1

19.81m/s

Step-by-step explanation:

To find the speed at which the car leaves the cliff, first we need to know how long it took to reach the ground. We do this with the equation:

`y=y_(0)-(1)/(2)gt^2`

Where
`y` is the vertical position at a time
`t`, since we are looking for the moment when the car reaches the gound:
`y=0`.
`y_(0)` is the initial vertical position, in this case the height of the cliff:
`y_(0)=125m`.
`g` is gravitational acceleration:
`g=9.81m/s^2`.

So replacing the known values:

`0=125-(1)/(2)(9.81m/s^2)t^2`

and since we need the time, we clear for it:

`0=125-(1)/(2)(9.81m/s^2)t^2\\-125=-(1)/(2)(9.81m/s^2)t^2\\(-125)(-2)=(9.81m/s^2)t^2\\(250)/(9.81m/s^2)=t^2 \\25.48=t^2\\√(25.48)=t\\ 5.048=t`

At time
`t=5.048` the car reaches the gound, according to the problem at a horizontal distance from the base of the cliff of 100m.

To find the velocity we use:

`x=x_(0)+v_(0)t`

where
`x` is the horizontal distance at time
`t` (in this case
`x=100` and
`t=5.048`),
`x_(0)` is the initial horizontal distance, we define the cliff as zero in horizontal distance so
`x_(0)=0`, and
`v_(0)` is the velocity of the car when it rolled of the cliff.

replacing the known values:

`100=0+v_(0)(5.048)`

and we clear for
`v_(0)`:

`(100)/(5.48) =v_(0)\\19.81m/s=v_(0)`

The initial velocity (when it rolled of the cliff) is 19.81m/s