How Does a Small Change in Speed and Angle Affect Projectile Range?

[reddawg]

Homework Statement

The range of a projectile fired (in a vacuum) with initial velocity V₀ and inclination angle θ from the horizontal is R = (1/32) * (V₀)² * sin(2θ).

Use differentials to approximate the change in range if V₀ is increased from 400 ft/s to 410 ft/s and θ is increased from 30° to 31°.

Homework Equations

Setting up my equation:

dR = (∂R/∂V₀) * dV₀ + (∂R/∂θ) * dθ

The Attempt at a Solution

Taking the partials and substituting values:

sin(60°)/250 + cos(60°)/10,000 = 5216.51 ft.

This answer seems unreasonable, can anyone check my work?

 

[SteamKing]

It's not clear what you have done since you don't show all the steps in taking your differentials.

Did you apply all of the derivative rules correctly? I can't tell from the attempt at your solution, since you have already substituted numbers. What happened to the factor (1/32) from the original range equation?

 

[reddawg]

The attempt at a solution:

Taking the partials:

∂R/∂V₀ = (sin(2θ)*V₀)/16

∂R/∂θ = (cos(2θ)*(V₀)²)/16

dV₀ = 10

dθ = 1

I then substituting and simplified.

 

[SteamKing]

Did you use radian measure for your dθ?

 

[reddawg]

I thought I had calculator in degree mode, however I went back through with radian and verified d(theta) was in radians.

Also there was a sign error in my calculations so the new answer is 129.24 ft which seems more reasonable don't you think?

 

[SteamKing]

Yes.