Question

reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?

Answer 1

Answer: θ would equal approximately 28.7°

This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.

Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°

Now if we multiply the range by 2, we get:

2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:

2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ

Thus, θ = 28.67780425

It's been awhile since I did this; though I hope it helped!