Question

Question #0a442

Answer 1

It hit ground again 338 m away.

Explanation:

There is a standard formula for range, R, that you can plug data into:
`R = (u^2*sin(2theta))/g`
There is one catch. You need to know the value of g on the moon. It is 1/6th of what it is on Earth, so `(9.8 m/s^2)/6 = 1.6 m/s^2`.

I think it would be more valuable to show a method that relies on an understanding of the physics.

The length of time it will be in the air is given by looking at the vertical motion using the suvat formula
`s = u*t + (a*t^2)/2`
Plugging our data

• s = 0 because it will return to the elevation it started with
• u = `(25 m/s)*sin30` because that is the vertical component of the velocity
• a = -1.6 m/s^2 that is the moon's value of g and I gave it a minus sign because I gave a plus sign to the initial velocity and g is downward

into that gives us
`0 = (25 m/s)*sin30*cancel(t) + (-1.6 m/s^2*tcancel(^2))/2`

`(1.6 m/s^2*t)/2 = (25 m/s)*sin30`

`t = 2*(25 m/s)*sin30/ (1.6 m/s^2) = 15.6 s`

How far will it travel horizontally during 15.6 s?
The horizontal velocity is `(25 m/s)*cos30 = 21.6 m/s`

The horizontal displacement is
`s = v*t = 21.6 m/s * 15.6 s = 338 m`

I hope this helps,
Steve