Determine the Optimal Projection Angle for Twice the Horizontal Range as Maximum Height

In the realm of projectile motion, finding the angle of projection that makes the horizontal range exactly twice the maximum height is an intriguing challenge. This article will guide you through the mathematical derivation to find this specific angle, providing a comprehensive understanding of projectile dynamics.

Equations of Motion for Projectile Motion

To solve for the angle of projection θ that satisfies the condition where the horizontal range is twice the maximum height, we can use the following equations derived from the laws of motion:

Maximum Height
Maximum Height H Hu22gsin2θ2 where u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

Horizontal Range
Horizontal Range R

Ru2gsin22θ

Given that the horizontal range is twice the maximum height, we set:

R2H

Substituting the expressions for R and H into the equation:

u2gsin22θ2u22gsin2θ2

After simplifying this equation, we get:

u2gsin22θu2gsin2θ2

Canceling out u2g from both sides, we get:

sin22θsin2θ2

Using the double angle identity sin 2θ 2 sin θ cos θ, we substitute and rearrange the equation:

sin22θsin2θ2 2sinθ2cosθ sinθ sinθsin2θ2

Factoring out sin θ, we get two cases:

Case 1: sin θ 0
This corresponds to θ 0° or 180°, which are not useful for projectile motion.

Case 2: sin θ - 2 cos θ 0
or sin θ 2 cos θ

Dividing both sides by cos θ gives us:

tanθ 2

Thus, the angle of projection θ is:

θ atan2 63.43°

Conclusion: The angle of projection at which the horizontal range is twice the maximum height is approximately 63.43 degrees.

References and Further Reading

For a deeper understanding of projectile motion, consider exploring the following resources:

Projectile Motion - Khan Academy Projectile Motion - PhysionWeb Projectile Motion - The Physics Classroom