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 F? wind=?Fwindi^.

By what percentage is the maximum range of a 0.900 kg ball reduced if Fwind =0.600 N ?

Express your answer with the appropriate units.

*Answer cannot be 7.25 %, 6.8%, or 9% nor 6.3%

Answer 1

a)

mass of the projectile is m

initial velocity of the projectile is vo

Horizontal distance travel is

X = vo*cos(theta) + 1/2at^2

X = vo*cos(theta) + 1/2*(F/m)t^2

t is the initial time of the flight which is equal to the time taken at which verticle displacement of the projetile is zero

0 = vo*sin(theta) + 1/2gt^2

t = 2vo*sin(theta) / g

X = (vo*cos(theta))*( 2vo*sin(theta) / g) - 1/2*(F/m)*( 2vo*sin(theta) / g)

X = (vo^2 / g)*sin(2*theta) - (2F/m)*(vo^2/g^2*sin^2(theta))

Now X be the maximum then dx/d(theta) = 0

0 = (vo^2/g)*sin(2*theta) - (2F/m)*(vo^2/g^2*sin^2(theta))

(vo^2/g)*sin(2*theta) = (2F/m)*(vo^2/g^2*sin^2(theta))

tan(2*theta) = mg/F

theta = 1/2*tan^-1(mg/F)

b)

m = 0.900 kg

Fwind = 0.600 N

theta = 1/2*tan^-1(0.900*9.81/0.600)

theta = 43 degree

X = (vo^2 / g)*sin(2*43) - (2F/mg)*(vo^2/g^2*sin^2(43))

X = vo^2/g*[sin(2*43) -  (2F/mg)**sin^2(43))

X =  vo^2/g*[sin(2*43) -  (2*0.600/0.900*9.81)*sin^2(43))

X = vo^2/g*[1.364 - 0.0632)

X = vo^2/g(1.3)

Xmax = vo^2/g

percentage of reduction in range is

(Xmax - x / Xmax)*100

[[vo^2/g - vo^2/g(1.3)] / vo^2/g ] x 100

sorry not sure about last one