What is the initial angle 0f projection of a projectile for which the range and the maximum height are equal?
To find the initial angle of projection for which the range and maximum height of a projectile are equal, we can use the following steps:
1. Let's assume the initial angle of projection as θ.
2. The range (R) of a projectile is given by the formula: R = (v^2 * sin(2θ)) / g,
where v is the initial velocity of the projectile and g is the acceleration due to gravity.
3. The maximum height (H) of a projectile is given by the formula: H = (v^2 * sin^2(θ)) / (2g).
4. Now, since we want the range and maximum height to be equal, we can set R = H.
5. Substituting the formulas for R and H, we get: (v^2 * sin(2θ)) / g = (v^2 * sin^2(θ)) / (2g).
6. We can simplify this equation by canceling out v^2 and g from both sides.
sin(2θ) = sin^2(θ) / 2.
7. Using the double-angle formula for sine (sin(2θ) = 2sin(θ)cos(θ)), we can rewrite the equation as:
2sin(θ)cos(θ) = sin^2(θ) / 2.
8. Multiply both sides of the equation by 2 to get rid of the fraction.
4sin(θ)cos(θ) = sin^2(θ).
9. Using the identity for double-angle sine (2sin(θ)cos(θ) = sin(2θ)), we can rewrite the equation as:
sin(2θ) = sin^2(θ).
10. Now we have an equation only in terms of sine.
11. To solve this equation, we can take the arcsine (inverse sine) of both sides:
2θ = arcsin(sin^2(θ)).
12. Finally, we can solve for θ by dividing both sides of the equation by 2:
θ = arcsin(sin^2(θ)) / 2.
Therefore, the initial angle of projection for which the range and maximum height of a projectile are equal is given by θ = arcsin(sin^2(θ)) / 2.