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Sonja [21]
3 years ago
12

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.
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0
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.
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This is my answer, and I'm stickin to it.

But ... just like all the other problems you get in high school, the
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I don't know how much effort you put into this problem, but somewhere
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I hope my work is clear enough that you can go through it, and pick up
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