Question

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon’s gravity is about 1/6th that of Earth.)

Answer 1

`x = (v_(0)) ^(2) * (sin 2 \alpha)/(g) + x_(0)`

If g goes down by a factor of 6, the distance x goes up by a factor of 6.

Answer 2

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth