In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an angle of 15 ∘ below the horizontal. It is released 0.80 m above the floor.

Question

Elliana Jacobs

Physics

5 May, 06:59

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an angle of 15 ∘ below the horizontal. It is released 0.80 m above the floor.

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Larry Cantrell · Answer 1

The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ + / - gx²/2v² (cosθ) ²

where

y is the vertical distance

x is the horizontal distance

θ is the angle of trajectory or launch angle

g is 9.81 m/s²

v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²) x²/2 (4.8 m/s) ² (cos15°) ²

Solving for x,

x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.