Motion Help Please

  • Thread starter Kotune
  • Start date
  • #1
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Homework Statement


A stunt car is driven from a 10° ramp at 40m/s. How far away should the landing ramp be. (It lands at the same level as it was launched from.)


Homework Equations


Answer is 54.7m


The Attempt at a Solution


v^2 = u^2 + 2as
(v^2 - u^2)/(2a) = s
(0 - 40^2) / (2 x 9.8) = -81.6
 

Answers and Replies

  • #2
537
1
try splitting the original velocity into x and y components. Remember that gravity only affects the velocity component in the y direction.
 
  • #3
u r right. gravity will affect only vertical component. horizontal component remains same.
 
  • #4
21
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try splitting the original velocity into x and y components. Remember that gravity only affects the velocity component in the y direction.

I don't understand. What equations am I suppose to use?
 
  • #5
537
1
Well, if the velocity of the stunt car is 40 m/s at 10 degrees initially, then how would you find the x and y components of its velocity at that moment?
 
  • #6
21
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Well, if the velocity of the stunt car is 40 m/s at 10 degrees initially, then how would you find the x and y components of its velocity at that moment?

x = 40cos10 = 39.39
y = 40sin10 = 6.95

I get that part but I'm not sure how that helps me achieve the conclusion? ><
 
  • #7
537
1
Well so if the car flies off into the air, how do you find out how long it's going to stay in the air?
 
  • #8
21
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x = 40cos10 = 39.39
y = 40sin10 = 6.95

I get that part but I'm not sure how that helps me achieve the conclusion? ><

Well so if the car flies off into the air, how do you find out how long it's going to stay in the air?

v = u + at?
0 = 6.95 +9.8t
-6.95 / 9.8 = t
0.7 = t
0.7s
 
  • #9
537
1
that's what I got. So now you know how long the car is in the air and how fast it is going in the x direction, how would you find out how far away it lands?
 
  • #10
21
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.7 x 39.39
= 27.573
Then do I multiply this number by 2?
 
  • #11
537
1
well, what do you think? ;)
 
  • #12
21
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got it, cheers buddy!!
 

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