Physics Problem with Variables and Projectile Motion

Question

So we have a physics problem where there is a basketball hoot 1.2 meters above the ground and someone shoots a ball from 10 meters away (x distance) at a 50 degree angle with the ground and we have to find the initial velocity (not specific x/y velocity) so that the ball will go into the hoop. We have to find an equation for initial velocity with only the variables g (gravity acceleration), D (x displacement), H (y displacement), and theta (angle) and the equation that my teacher simplified it down to was Vo = square root of ( (g* D^2)/((2cos^2theta)(tantheta*D-H)) ) (the square root covers all of the right side). He told us for our test we need to simplify that equation and plug in those numbers to find the initial velocity. How can I simplify this and solve for Vo?

Answer 1

Define

D = 10 m, be the distance from the ball shooter to the basketball hoop;

H = 12 m, the height of the hoop above ground;

theta = 50 deg., the launch angle, measured above the horizontal;

V₀ = launch velocity, to be determined;

g = 9.8 m/s², acceleration due to gravity.

Ignore air resistance.

The horizontal component of the velocity is V₀ cos(theta).

The vertical component of the velocity is V₀ sin(theta).

The time, t, for the ball to travel the horizontal distance, D, is

t = D/[V₀cos(theta)]. (1)

Within the same time, the ball should ascend to a height of H.

Use the formula s = ut +(1/2)at².

For our problem, s=H, a = -g.

Therefore

H = V₀ sin(theta)*t - (g/2)*t² (2)

Substitute equation (1) into (2).

H = [DV₀ sin(theta)] / [V₀ cos(theta)] - (gD²) / (2V₀² cos²(theta))

H = D tan(theta) - (gD²) / (2V₀² cos²(theta))

Rearrange to obtain

(gD²)/(2V₀² cos²(theta)) = D tan(theta) - H

Take the reciprocal of each side.

(2V₀² cos²(theta)) / (gD²) = 1/(D tan(theta) - H)

Multiply each side by (gD²)/(2 cos²(theta)).

V₀² = (gD²) / [2 cos²(theta) (D tan(theta) - H)]

Take the square root of each side to obtain the required answer.

V₀ = [(gD²) / {2 cos²(theta) (D tan(theta) - H)} ]^1/2