Any help on how to solve this?
The range of a projectile launched at an angle θ from the ground with velocity v0 is given by R(θ) = [v0^2 sin 2(θ)]/9.81 . If the projectile is launched at an angle of θ = π/6, use differentials to approximate the percentage change in the range of the
projectile if the angle was increased by 2%.
using T for theta
dR/dT = (Vo^2 /9.81) cos2T
(note max range when T = 45 degrees :) )
100 dT/T = 2
dT = .02 T = .02 * pi/6
dR = (Vo^2 /9.81) cos2T dT
100 dR/R = cos 2T dT / sin 2T
To find the percentage change in the range of the projectile, we need to calculate the differential of the range with respect to the angle and use it to approximate the change.
1. Start by differentiating the range equation with respect to θ:
dR(θ) = d([v0^2 sin 2(θ)]/9.81)
2. Apply the chain rule to differentiate sin 2(θ):
dR(θ) = [v0^2 cos 2(θ)]/4.905 dθ
3. Substitute the given angle θ = π/6 and its change of 2% (in radians) into the differential equation:
dR(π/6) = [v0^2 cos 2(π/6)]/4.905 (0.02)
4. Calculate the new range R_new by adding the differential to the original range R_old:
R_new = R(π/6) + dR(π/6)
5. Calculate the percentage change in range:
Percentage Change = [(R_new - R_old)/R_old] * 100
By following these steps, you should be able to approximate the percentage change in the range of the projectile.