A daredevil jumps a canyon 13 m wide. To do so, he drives a car up a 13 ◦ incline.

What minimum speed must he achieve to clear the canyon?
The acceleration of gravity is 9.81 m/s^2.
Answer in units of m/s

If the daredevil jumps at this minimum speed, what will his speed be when he reaches the other side?
Answer in units of m/s

use the range equation

R = v0^2 sin(2 theta)/g

R = 13m
theta = 13 deg
g=9.81m/s/s

v0 = Sqrt[g R/sin(2 theta)] = Sqrt[9.8 x 13/sin 26] = 17.0m/s

on reaching the other side, the speed will be the same as at liftoff, assuming no air friction

OVER 9000 m/s