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...

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.