Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45∘ angle. Suppose a projectile of mass m is launched with speed v0 into a headwind that exerts a constant, horizontal retarding force F⃗ wind =−Fwindi^.

By what percentage is the maximum range of a 0.600 kg ball reduced if Fwind = 0.100 N ?

Express your answer with the appropriate units.

Answer 1

Range of a projectile is given by

R = (v₀²sin2)/g

For angle of projection = 45°

R = ( v₀²sin90°)/g

R = v₀²/g

When the wind starts blowing, retardation of the projectile

a_x = -F/m = -0.1/0.6 = -1/6

Time of flight T = (2v₀​​​​​​sin45°)/g remains unchanged as there is no change in the acceleration along the y axis

For horizontal motion

S_x = u_xT + (1/2)a_xT²

Let S_x = R' = New horizontal range of the projectile

u_x = v₀​​​​​​cos45° = initial horizontal velocity

T = time of flight = 2v₀​​​​​​sin45°/g

R' = v₀cos45°×2v₀sin45°/g - (1/2)×(1/6)×(2v₀sin45°/g)²

R' = v₀²/g - (1/12)×(2v₀²/g²)

R' = v₀²/g - (1/6)v₀²/g²

Percentage change in the range is given by

(R' - R)×100/R = [ ( v₀²/g -v₀²/6g²) - (v₀²/g) ]×100 / (v₀²/g)

= - 100/6g

= - 100/60

= - 1.67 %

Negative sign indicates that range is decreasing.

Therefore range of the projectile decreases by 1.67 %