# Calculate the angle of projection, with the horizontal, for a projectile for which the maximum height is equal to the horizontal range

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The angle of projection for which the maximum height of the projectile is equal to the horizontal range has to be determined.

Let the required angle of projection be A and object is projected at V m/s.

The horizontal component of the velocity is V*cos A and the vertical component is V*sin A. There is only an acceleration equal to g acting vertically downwards on the object. The horizontal component of the velocity does not change.

At the maximum height the vertical velocity is equal to 0. This happens at time t where V*sin A - g*t = 0

=> t = V*sin A/g

The particle travels for a time t = 2*V*sin A/g and the horizontal range is equal to (2*V*sin A/g)*V*cos A

The maximum height is V*sin A*V*sin A/g - (1/2)*g*(V*sin A/g)^2

=> (V*sin A)^2/g - (1/2)*(V*sin A)^2/g = (1/2)*(V*sin A)^2/g

As this is equal to the range:

(1/2)*(V*sin A)^2/g = (2*V*sin A/g)*V*cos A

=> (1/2)*sin A = 2*cos A

=> sin A/cos A = 4

=> tan A = 4

=> A = 75.96 degrees

**If the angle of projection is 75.96 degrees the maximum height is equal to the horizontal range.**