A car drives straight off the edge of a cliff that is 58 m high. The police at the scene of the accident note that the point of impact is 0.136 km from the base of the cliff. How long did it take to hit the ground?
It took the car
Here's what's going on here.
Because the car drives straight off the cliff, you know for a fact that the vertical component of its initial velocity is equal to zero.
Once the car drives off the cliff, its motion will be affected by the gravitational acceleration.
In fact, if you break down its motion on two dimensions, you can say that vertically the car is free falling because the vertical component of its initial velocity is equal to zero.
Horizontally, however, the car is not acted upon by any force, which implies that horizontal component of its initial velocity is constant.
So, if you take
#d = underbrace(v_(0x))_(color(blue)(=v_0)) * t ->#horizontally
#h = underbrace(v_(oy))_(color(blue)(=0)) * t + 1/2 * g * t^2 ->#vertically
Since you don't know the speed with which the car leaves the cliff, which I called
#h = 1/2 * g * t^2 implies t = sqrt((2h)/g)#
#t = sqrt( (2 * 58color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^2)) = color(green)("3.44 s")#