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 level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
can anyone help me?

Question

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 level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
can anyone help me?

john_es · Answer

Time to reach to the ground,
$$0=50-\frac{1}{2}gt^2\Rightarrow t=3.19\ s$$So,
$$x=v_0t\Rightarrow v_0=x/t=90/3.19=28.2\ m/s$$