In a basketball game, point guard A intends the speed as 47.76 ft/sec, and the launch angle will be 41.084°.
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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