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