Linear Approximation of Angles

[Onodeyja]

Homework Statement

A player located 18.1 ft from a basket launches a successful jump shot from a height of 10 ft (level with the rim of the basket), at an angle Θ = 34 degrees and initial velocity of v = 25 ft/s.A. Show that the distance s of the shot changes by approximately 0.255∆Θ ft if the angle changes by an amount ∆Θ. Remember to convert the angle to radians in the Linear Approximation.

B. Is it likely that the shot would have been successful if the angle were off by 2 degrees?

Homework Equations

s = (1/32)v² * sin(2Θ)

The Attempt at a Solution

I'm not sure if I'm going in the right direction, but here's what I have.

∆f = f'(a)∆Θ

f' = 1250/32 * cos(2Θ)

∆f = f'(34 * (∏/180)) * ∆Θ
∆f = 1250/32 * cos(2 * (17∏/90)) * ∆Θ
∆f = 1250/32 * cos (17∏/90) * ∆Θ
∆f = 32.38∆Θ = 0.255
∆Θ = 0.255/32.38 = 0.00787 * (180/∏) = 0.45˚

Part B: No, the shot wouldn't be successful.

 

[Kreizhn]

Your statement
∆f = f'(a)∆Θ
Certainly seems to be correct, modulo notation. In particular, you meant f=s, a =\theta right?

After that, your calculations looks fine.

Also, what is your motivation for answering "No" to part b? Can you numerically justify it?

 

[Onodeyja]

Your statement
∆f = f'(a)∆Θ
Certainly seems to be correct, modulo notation. In particular, you meant f=s, a =\theta right?

After that, your calculations looks fine.

Also, what is your motivation for answering "No" to part b? Can you numerically justify it?

Yes, I should have written the function as s instead of f.

Θ = 34˚ = 17∏/90
∆Θ = 2˚ = ∏/90

∆s = s'(Θ)∆Θ = s'(17∏/90)(∏/90)
∆s = [1250/32 * cos(17∏/90)] * (∏/90)
∆s = 1.13

18.1 ft + 1.13 ft = 19.23 ft