Physics I: Projectile motion on a ramp.

Here's a good one.

In one contest at the country fair, a spring-loaded plunger launches a ball at a speed of 3.0 m/s from one corner of a smooth, flat board that is tilted up at a 20 degree angle. To win, you must make the ball hit a small target at the adjacent corner, 2.50 m away. At what angle should you tilt the ball launcher?

The ball is rolling up and down the ramp, from one corner to another. This really isn't much different from a standard projectile motion problem, the only difference is the y component acceleration is -gsin20 rather than simply -g.

A few things to note: we have the initial velocity but not its components, so this is quite do-able. Be careful not to mix the slope of the ramp and the unknown angle we're trying to find!

v_0x = v₀cosθ
v_0y = v₀sinθ 

Solve for t when v_y = 0 (the peak of the trajectory).

v_y = v_0y + a_yt = v₀sinθ - gtsin20
gtsin20 = v₀sinθ (plugging in v₀sinθ for v_0y)
t = (v₀sinθ)/gsin20

Remember, this is the time when the ball is at its highest point, make sure not to forget that we're using x₁ (2.50 m / 2), not x₂ (the final position).

x = v₀t
x₁ = v_0xt₁
x₁ = 2v_0xt₁
x₁ = v_0x(v₀sinθ/gsin20)
x₁ = v₀²cosθsinθ/gsin20 (plugging in v₀cosθ for v_0x)
x₁ = v₀²sin2θ/2gsin20 (using double angle formula 2sinθcosθ = sin2θ, since we didn't have a 2 I multiplied the whole fraction by 2/2)
sin2θ = 2_x1gsin20/v₀²
sin^-1(2_x1gsin20/ v₀²) = 2θ (reminder, plug in 1.25m for x₁, not 2.50, since this is x at t₁ which is at ½ t_f)
2θ = 68.6 degrees

Solution: θ = 34.3 degrees