The angle of projection at which the horizontal range and maximum height of projectile are equal is

A) 45

B) θ=tan−1(0.25)

C) θ=tan−14orθ=75.96

D) 60

To find the angle of projection at which the horizontal range and maximum height of a projectile are equal, we can use the following formula for the horizontal range (R) and the formula for the maximum height (H) of a projectile:

Horizontal Range (R):
R = (u^2 * sin(2θ)) / g

Maximum Height (H):
H = (u^2 * sin^2(θ)) / (2g)

Where:

R is the horizontal range.
u is the initial velocity of the projectile.
θ is the angle of projection.
g is the acceleration due to gravity (approximately 9.81 m/s²).
To find the angle of projection at which R = H, we can set these two equations equal to each other:

(u^2 * sin(2θ)) / g = (u^2 * sin^2(θ)) / (2g)

Now, let's cancel out u^2 and g from both sides of the equation:

sin(2θ) = (sin^2(θ)) / 2

Now, let's solve for θ. Start by substituting the double angle formula for sin(2θ):

2sin(θ)cos(θ) = (sin^2(θ)) / 2

Now, cancel out the common factor of 2:

sin(θ)cos(θ) = (sin^2(θ)) / 4

Divide both sides by sin(θ) to isolate cos(θ):

cos(θ) = (sin(θ) / 4)

Now, divide both sides by sin(θ):

cos(θ) / sin(θ) = 1 / 4

Using the identity tan(θ) = sin(θ) / cos(θ), we can rewrite the equation:

tan(θ) = 1 / 4

Now, to find θ, take the arctan (inverse tangent) of both sides:

θ = arctan(1 / 4)

Now, calculate the arctan:

θ ≈ 14.04 degrees

So, the angle of projection at which the horizontal range and maximum height of the projectile are equal is approximately 14.04 degrees.

None of the answer choices exactly matches this result, but the closest one is:

C) θ = tan^(-1)(1/4) or θ ≈ 14.04 degrees

Therefore, option C is the closest choice to the correct answer.