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.

giorgiana1976 | Student

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.