Question

An object released at an angle of inclination θ with initial velocity v0 ​ft/s travels a horizontal distance s=1/32 v0²​sin(2θ)ft. A basketball player located 18.1ft from the basket launches a successful jump shot from a height of 10ft (level with the rim of the basket), at an angle θ=34∘ with initial velocity v0​=25ft/s. Find the differential ds.

Answer 1

The differential ds is the change in the horizontal distance with respect to the change in the launch angle θ, and for v0 = 25 ft/s and θ = 34° it's approximately 14.64 ft/deg. The differentiation is performed applying the chain rule to the given formula for s.

Step-by-step explanation:

The student asks about the horizontal distance s traveled by an object with an initial velocity v0 at an angle of inclination θ.

To find the differential ds, we'll differentiate the given formula s = 1/32 v0² sin(2θ) with respect to θ, because we're interested in how the horizontal distance changes with a small change in the angle of launch.

To perform the differentiation:

1. Identify the variables: s (horizontal distance), v0 (initial velocity), and θ (launch angle).
2. The initial velocity v0 is constant, so treat it as such during differentiation.
3. Differentiate with respect to θ using the chain rule since we have a sine function of 2θ.

Perform the differentiation:

ds/dθ = 1/32 × 2v0² cos(2θ)

Plugging in the given values of v0 = 25 ft/s and θ = 34°:

ds/dθ = 1/32 × 2(25)2 cos(2· 34°)
= 1/32 × 1250 × cos(68°)
= 39.0625 × cos(68°)
= 39.0625 × 0.3746
= 14.64 ft/deg

The differential ds is approximately 14.64 ft/deg, which means the horizontal distance traveled will change by about 14.64 feet for each degree of change in the angle θ.