# Projectile motion over angled edge!

1. May 16, 2012

### Samyuerux

I am having difficulties with this question as no height for the crater is given please help! :)

1. The problem statement, all variables and given/known data
AN astronaut is speeding over a level stretch of the Moon's surface at 12m/s in her small lunar rover. She drives over the edge of a crater, whose surface slopes downwards at an angle of 30 degrees below the horizontal.

1) It is known that g = 2m/s^2 on the Moon. How far down the slope (measured from the lip of the crater) will the vehicle land?

2) What is the velocity and speed of the vehicle before landing?

2. Relevant equations
S(t) = a/2t^2 + V(0)t + S(0), where V = velocity , a = acceleration and s = displacement.
Vf = Vi + at

3. The attempt at a solution

So i think the distance down the slope would be 6.93 Metres but im not sure if this is right, i used basic trig to work this out, then the rest i can work out myself just need to check that this is right or not!

Last edited: May 16, 2012
2. May 16, 2012

### tiny-tim

Welcome to PF!

Hi Samyuerux! Welcome to PF!
The ground is horizontal, then sloping down at 30° …

the projectile starts horizontally, and follows a parabola until it hits the slope …

how far down is that?

3. May 16, 2012

### Samyuerux

IM not sure how you work it out ?

Last edited: May 16, 2012
4. May 16, 2012

### Samyuerux

is it 6.93 Metres?
Fast reply would be great thank you!

5. May 16, 2012

### tiny-tim

i've no idea

if you want us to check your work, you need to show it!

6. May 16, 2012

### Samyuerux

Well i set up a basic right angled triangle with the Hyp being the slope downwards at 30 degrees.
The horizontal would then be set at 12.
this gives me a vertical componant thats 6.93, so i presumed that would be the distance but i think thats wrong and there is a more complicated method of finding it out which i dont know about.

7. May 16, 2012

### tiny-tim

no, you need to do standard constant acceleration equations for the x and y directions, separately (against t)

then use y = xtan30° to solve them …

what do you get?

8. May 16, 2012

### Samyuerux

I think i get you now,
so far i have Sy = t^2 and Sx = 12t
so i think i set Sy = t^2 as xtan30 = t^2

9. May 16, 2012

### Samyuerux

so then i rearrange to get 12 tan 30 = t
so t = 6.93
and then times t by Vx = 83.16 and t by Vy = 13.86
then do pythag to find the total displacement?
which comes out at 84.3
that seems too high...

10. May 16, 2012

### tiny-tim

looks ok

this is on the moon, remember!

11. May 16, 2012

### Steely Dan

Not quite. You have it correct for the $x$ distance ($x = v_x t$), but the vertical distance $y$ still satisfies the relationship $y = t^2$, and that is what you want to use, since the vertical velocity is not constant in time.

By the way, in the future, please post projectile motion problems in the "Introductory Physics" forum -- believe it or not, it's more likely to get answered there!

12. May 17, 2012

### azizlwl

Rsin60°=12t
Rcos60°=1/2at2

t=(Rsin60°)/12

Rcos60°=0.5a( (Rsin60°)/12 )2
R=(12.12.cos60°)/sin60°.sin60°
R=96m.

t=6.93sec.

Last edited: May 17, 2012
13. May 17, 2012

### azizlwl

x=83.16m
y=(1/2)at2=(6.93)2=48.02m

14. May 18, 2012

### Samyuerux

Thanks very much guys got their in the end! :)