What is the launch angle such that the maximum height of the projectile is equal to its horizontal range.The initial speed the projectile is launched is of 54.2 m/s.Ignore air resistance.
We'll recall the basic motion equation applied to determine the horizontal range:
x = v0*t => t = v0/x
Since the maximum height is equal to the horizontal range, we'll get:
v0x = v0y
(v0^2*sin 2a)/g = (v0^2)*(sin a)^2/2g
We'll recall the double angle identity:
sin 2a = 2 sin a*cos a
We'll simplify both sides:
sin 2a = (sin a)^2/2
2 sin a*cos a = (sin a)^2/2
4sin a*cos a = (sin a)^2
We'll simplify by sin a both sides and we'll get:
4 cos a = sin a
We'll divide by cos a:
sin a/cos a = 4 => tan a = 4 => a = arctan 4 => a = 76 degrees approx.
The launch angle of the projectile, under the given constraints, is: a = 76 degrees approx.