How fast must the motorcycle leave the cliff top to land

Question

A movie stunt driver on a motorcycle speeds horizontally off a 70.3 m high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 49. Percent m from the base of the cliff where the cameras are? Ignore air resistance

Answer 1

I'll assume you meant that the motorcycle lands 49 m from the base of the cliff.

Thinking ONLY about what happens in the vertical direction, There is no initial vertical velocity (the motorcycle takes off from the edge of the cliff going only in the horizontal direction. Therefore, from a vertical perspective this is just like free fall and y = 1/2gt² or t = √(2y/g) = √(2•70.3/9.81) = 3.786 seconds.

Now, thinking ONLY in the horizontal direction, the motorcycle must travel 49 m in 3.786 seconds so it must be traveling 12.94 m/s. If we assume 2 significant figures, that rounds to 13 m/s