How do I solve this

Question

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of  54 degrees, the crew fires the shell at a muzzle velocity of 174 ft/s. How far (d) meters down the hill does the shell strike if the hill subtends an angle 36 degrees from the horizontal? Ignore air friction. How long will the shell be in the air (t) seconds. Then how fast will the shell be traveling when it hits the ground? (speed) m/s

Answer 1

Let's set origin at the top of the hill with y up and x to the right

For the projectile: x(t) = v_0xt and y(t) = v_0yt - 1/2 gt² where v_0x = v₀cos(θ₀) and v_0y = v₀sin(θ₀)

The hillside has the equation y/x = -tan(φ) or y = -xtan(φ) (here φ is 36°)

You can make the 2nd equation in terms of t: y = -v_0xt tan(φ)

To find time in the air: equate the y(t) and solve the quadratic for t

To find d, plug into the equations for y and x at t_air and find d = sqrt(x² + y²)

To find v(t_air) : use the Pythagorean theorem to combine v_x = v_x0 and v_y = v_y0 - gt_air

Serious problem!

PS. use 32.17 ft/s² or go metric throughout.