Question

If a modern day Robin Hood wanted to shoot his bow on the moon, how far could the arrow go? Assume that the bow acts like a spring that can store a total of 100 J. The mass of each arrow is 0.1 kg. The mass of moon is 7.34767309 × 1022 kg and its radius is 1737.4 km. The gravitational constant G = 6.673×10-11 N m2 kg-2. Also, what would the range for each arrow be if Robin fired six of them at the same time? Hint: note that cos and sin for an angle can only have values between -1 and 1.

Answer 1

acceleration due to gravity on the moon,

g = G*M/R^2

= 6.673*10^-11*7.34767309*10^22/(1737.4*10^3)^2

= 1.624 m/s^2

let v is the speed of the arrows.

the kinetic energy of the 6 arrows is equal to 100 J .

KE = 6*(1/2)*m*v^2

KE = 3*m*v^2

v = sqrt(KE/(3*m))

= sqrt(100/(3*0.1))

= 18.26 m/s

for maximum range angle projection must be 45 degrees.

so,

R_max = v^2*sin(2*theta)/g

= 18.26^2*sin(2*45)/1.624

= 205.3 m <<<<<<<<<-----------------Answer