The highest barrier that a projectile can clear is 13.7 m, when the projectile is launched at an angle of 10.5° above the horizontal. What is the projectile's launch speed?

Answer 1

You are going to work this problem kind of backwards. Meaning typically you have your launch speed and you have to break it down to find the vertical and horizonal velocity components before you can move forward to find lets say max height. In this case you are given max height and you know that your velocity in the vertical is zero at max height.

Solve for your initial velocity in the vertical (y direction) using the equation that does have time.

V_fy² = Viy² + 2gy so 0 = V_iy² + 2(-9.81)(13.7)...don't forget to square root it get final answer for initial velocity in the vertical (V_{i y})

Now go back to the triangle of your launch speed. Cos (theta) = (V_{i y})/V...V is the launch speed.