A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top to land on a level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

The object here is moving horizontally therefore the vertical component is zero. We first determine the time it takes for the object to move down the cliff bu using the equation,

d = (v1) (t) + (0.5) gt²

Substituting the given values,

50 m = 0 + (0.5) (9.8) (t²)

t = 2.26 s

For the horizontal component, the velocity is constant, thus:

90 m = (v) (2.26 s)

The velocity is 39.82 m/s.

Since the motorcycle moves horizontally, the vertical component of its velocity is zero. Determine the time it takes for the motorcycle to move down the 50-m cliff.

d = (v1) (t) + (0.5) gt²

Substituting the known values,

50 m = 0 + (0.5) (9.8) (t²); t = 2.26 s

Now for the horizontal component, the velocity is constant

90 m = (v) (2.26 s)

The value of v is approximately 39.82 m/s.