Question

A basketball is shot from an initial height of 2.40 m (Figure 1) with an initial speed v₀ = 14 m/s directed at an angle θ₀ = 30° above the horizontal. What is the projectile's range?

a) 12.8 m
b) 18.2 m
c) 24.5 m
d) 30.1 m

Answer 1

The range of the projectile is approximately 12.8 meters.

Step-by-step explanation:

To find the projectile's range, we can use the equations of projectile motion. We can break the initial velocity into its horizontal and vertical components. The horizontal component is given by v₀cos(θ₀) and the vertical component is given by v₀sin(θ₀).

Since there is no vertical acceleration (ignoring air resistance), we can use the equation Δy = v₀yt + 0.5gt² to find the time it takes for the projectile to hit the ground, where Δy is the initial height and g is the acceleration due to gravity. Plugging in the values, we get:

2.40 = (v₀sin(30°))t + 0.5(9.8)t²

Solving for t, we get t ≈ 0.55 seconds.

Now, we can find the horizontal distance traveled using the equation x = v₀xt:

x = (v₀cos(30°))(0.55)

Plugging in the values, we get x ≈ 12.8 meters.