A daredevil is shot out of a cannon at 33.7◦ to the horizontal with an initial speed of 25.5 m/s. A net is positioned at a horizontal distance of 43.7 m from the cannon from which the daredevil is shot.
The acceleration of gravity is 9.81 m/s^2
At what height above the cannon’s mouth should the net be placed in order to catch the daredevil?
Answer in units of m
Given: θ = 33.7; vo = 25.5 m/s; dx = 43.7m; a = 9.81m/s^2
vox = vo(cosθ)
voy = vo(sinθ)
[I'm stuck on what other equations I can use]
The Attempt at a Solution
I first drew a diagram with no problem.
vox = (25.5m/s)cos33.7 = 21.1m/s @ a = 0m/s^2
voy = (25.5m/2)sin33.7 = 14.1m/s @ a = -9.81m/s^2
So we now know initial velocity in the x direction is 21.1m/s, and initial velocity in the y direction is 14.1m/s. I am a little confused on where to go from here.