Ahobby rocket reaches a height of 72.3 meters and lands 111 meters from the launch point
what was the angle of launch?

Using the following formulas for projectile motion:

Height, H = ( Vo^2 * sin theta^2 )/g

Range, R = ( Vo^2 * sin 2*theta )/g

Rearranging in terms of Vo^2:

Vo^2 = gH / sin theta^2

Vo^2 = gR / sin 2*theta

Equating the two formulas to each other to solve for the angle theta:

gR / sin 2*theta = gH / sin theta^2

Substituting the given values:

(9.8)(111) / sin 2*theta = (9.8)(72.3) / sin theta^2
angle = 52.36 degrees

Therefore, the angle of launch is approximately 52.36 degrees.