Projectile motion over angled edge!

  • Thread starter Samyuerux
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  • #1
Samyuerux
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I am having difficulties with this question as no height for the crater is given please help! :)

Homework Statement


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?

Homework Equations


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

Thank you for your help!


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:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Samyuerux! Welcome to PF! :smile:
I am having difficulties with this question as no height for the crater is given

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? :wink:
 
  • #3
Samyuerux
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IM not sure how you work it out ?
 
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  • #4
Samyuerux
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is it 6.93 Metres?
Fast reply would be great thank you!
 
  • #5
tiny-tim
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is it 6.93 Metres?

i've no idea :confused:

if you want us to check your work, you need to show it! :wink:
 
  • #6
Samyuerux
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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
tiny-tim
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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? :smile:
 
  • #8
Samyuerux
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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
Samyuerux
10
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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
tiny-tim
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looks ok :smile:

this is on the moon, remember! :wink:
 
  • #11
Steely Dan
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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...

Not quite. You have it correct for the [itex]x[/itex] distance ([itex]x = v_x t[/itex]), but the vertical distance [itex]y[/itex] still satisfies the relationship [itex]y = t^2[/itex], 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
azizlwl
1,065
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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.
 
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  • #13
azizlwl
1,065
10
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...

x=83.16m
y=(1/2)at2=(6.93)2=48.02m
 
  • #14
Samyuerux
10
0
Thanks very much guys got their in the end! :)
 

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