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At what angle will the range of projectile be equal to half of maximum height of projectile?

User Mauro Dias
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Answer:

15 degrees

Step-by-step explanation:

Let's assume that the initial velocity of the projectile is v0, the angle of projection is θ, and g is the acceleration due to gravity.

The maximum height reached by the projectile can be calculated using the formula:

H = (v0² sin²θ)/(2g)

The range of the projectile can be calculated using the formula:

R = (v0² sin2θ)/g

We want to find the angle θ at which the range is half the maximum height, i.e., R = H/2. Substituting the equations for H and R, we get:

(v0² sin²θ)/(2g) = (v0² sin2θ)/2g * (1/2)

sinθ cosθ = 1/4

Now, we can use the identity sin2θ = 2 sinθ cosθ to write:

sin2θ = 2 sinθ cosθ = 2 * (1/4) = 1/2

Taking the inverse sine of both sides, we get:

2θ = 30°

θ = 15°

Therefore, the angle at which the range of the projectile is equal to half the maximum height is 15 degrees.

I hope this helped :)

User Nick Gowdy
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