In a basketball game, the point guard A intends to
throw a pass to the shooting guard B, who is breaking
toward the basket at a constant speed of 12 ft /sec. If
the shooting guard is to catch the ball at a height
of 7 ft at C while in full stride to execute a layup,
determine the speed v0 and launch angle with
which the point guard should throw the ball

The projectile motion is given as the motion of the object in the curved path near the earth's surface. The throwing of the ball will follow the projectile motion.

The missing image is attached below.

The given game situation consists of:

Velocity of shooting guard (vs) = 12 ft/sec.

Distance from B to C (sc) = 20 ft

Final height to catch the ball, 7 ft

The total distance to cover from B to C = 60 ft

Time is taken to reach B to C = Distance/Speed

Time to reach B to C = 20ft/12ft/sec

Time to reach B to C = 1.667 sec.

The time taken to cover the horizontal distance:

Distance = Velocity * Time

60³ = v₀cosθ * distance/time

60³ = v₀cosθ * 20/12

v₀cosθ = 36 ft/sec

The vertical movement can be given where, g = 32.2 ft/sec²:

s = s₀ + v₀sinθ * time + 1/2 -g time²

7 = s + v₀sinθ * 20/12 - 1/2 32.2 ft/sec² * (20/12)²

v₀sinθ = 31.3875 ft/sec

[tex]\rm v_0 =\sqrt{v_0sin\theta^2\;+\;v_ocos\theta^2}[/tex]

Substituting the values:

v₀ = 47.76 ft/sec

The launch angle for the ball will be:

tanθ = v₀sinθ/v₀cosθ

tanθ = 31.3875/36

tanθ = 0.871

θ = 41.084°

Thus, the speed v₀ of the ball will be 47.76 ft/sec, and the launch angle will be 41.084°.

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