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Fundamentals Of Physics
Found in: Page 90

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Short Answer

At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angleθ0=55°above the horizontal, to make the foul shot? The horizontal distances are d1=1.0ft and d2=14 ft, and the heights are h1=7.0 ft and h2=10.0ft.

The initial speed of the basketball is 23 ft/s

See the step by step solution

Step by Step Solution

Step 1: The given data

a) The angle made by the basketball above the horizontal, θ=55°b) The horizontal distances, d1=1 ft, d2=14 ftc) The heights are,h1=7 ft, h2=10 ft

Step 2: Understanding the concept of the projectile motion

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration. (Upward is taken to be a positive direction.)

Using the equation for the projectile path in the projectile motion of the ball, we can find the initial speed of the basketball.


The equation of the distance for the projectile path of a body,

y=x tanθ0-gx22V0cosθ02 …(i)

Step 3: Calculation of the initial speed of the basketball

The total horizontal displacement of basketball is given as:

x=d2-d1 =13 ft

The total vertical displacement of basketball is given as:

y=h2-h1 =3 ft

Rearranging the equation (i)for projectile path and substituting the above values, the velocity is given as:

V0=xcosθg2x tanθ-y =13 ftcos55° 32.15 ft/s2213 fttan55°-3 ft =13 ft0.57432 ft/s231.13 ft =22.98 ft/s 23 ft/s

Hence, the value of the initial speed of the ball is 23ft/s.

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