# Motion Help Please

## 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.)

## 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

try splitting the original velocity into x and y components. Remember that gravity only affects the velocity component in the y direction.

u r right. gravity will affect only vertical component. horizontal component remains same.

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?

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?

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? ><

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?

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

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?

.7 x 39.39
= 27.573
Then do I multiply this number by 2?

well, what do you think? ;)

got it, cheers buddy!!