At what projection angle will the range of a projectile equal to 5 times its maximum height?
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To find the projection angle at which the range of a projectile is equal to 5 times its maximum height, we need to break it down into steps. Let's go through the steps together:
Step 1: Identify the necessary variables.
- In this case, we need to know the range and maximum height of the projectile.
- The range is the horizontal distance traveled by the projectile.
- The maximum height is the highest point reached by the projectile.
Step 2: Determine the equations.
- The range equation for a projectile at an angle θ (theta) is given by: R = (v^2 * sin(2θ)) / g, where v is the initial velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The maximum height equation for the same projectile is given by: H = (v^2 * (sin(θ))^2) / (2g).
Step 3: Set up the equation.
- We are looking for the projection angle (θ) where the range equals 5 times the maximum height. So, we can set up the following equation: R = 5H.
Step 4: Substitute the equations and solve.
- Substitute the range equation (R) and maximum height equation (H) into the equation R = 5H.
- (v^2 * sin(2θ)) / g = 5 * (v^2 * (sin(θ))^2) / (2g).
- Simplifying, we get: 2sin(2θ) = 5sin^2(θ).
Step 5: Solve for θ.
- Rearrange the equation: 2sin(2θ) - 5sin^2(θ) = 0.
- Apply trigonometric identities: 2sin(θ)cos(θ) - 5sin^2(θ) = 0.
- Factoring out sin(θ): sin(θ) * (2cos(θ) - 5sin(θ)) = 0.
- Set each factor equal to zero: sin(θ) = 0 or 2cos(θ) - 5sin(θ) = 0.
Step 6: Solve for the projection angles.
- For sin(θ) = 0, the possible solutions for θ are θ = 0° (or multiples of 180°). This corresponds to a vertical projection, which means the projectile won't go anywhere.
- For 2cos(θ) - 5sin(θ) = 0, we can use the trigonometric identity: cos(θ) = √(1 - sin^2(θ)). Substituting, we get: 2√(1 - sin^2(θ)) - 5sin(θ) = 0.
- Simplifying, we have: 2√(1 - (sin(θ))^2) = 5sin(θ).
- Squaring both sides, we get: 4 - 4sin^2(θ) = 25sin^2(θ).
- Rearranging and simplifying, we have: 29sin^2(θ) = 4.
- Solving for sin(θ), we get: sin(θ) = √(4/29) ≈ 0.387.
- Taking the inverse sine (sin^-1) of both sides, we find: θ ≈ 22.8°.
So, the projection angle at which the range of a projectile equals 5 times its maximum height is approximately 22.8°.