Calculating Final Velocity of a Car Over a Ramp

[Drakon25th]

OK, many people in my class got a different answer from each other, so could someone verify my work for me please? Here's the problem:

A car (m=1000kg) driving at 20 m/s goes over a ramp angled at 37 degrees with a height of 5 m. How fast is the car traveling when it hits the ground 15 m below the end of the ramp?

my work:

Vyo = 20m/s ( sin37º)
Vyo = 12.0 m/s

Vy = sqrt(vyo^2 + 2gy)
Vy = sqrt ((12m/2)^2 + 2 * 9.8m/s^2 * 15 m)
Vy = 21 m/s

So my answer was 21 m/s

edit: my webspace is down right now so can't post pic

is this correct or am i doing something wrong?

 

[Warr]

You forgot something

When the car hits the ground, it is not traveling perfectly vertical. Remember, for projectile motion, the horizonal velocity remains constant. Therefore you must add the final vertical velocity and the final horizontal velocity together (note they are vectors) to get a resultant velocity, which is on an angle that you must also calculate based on these velocities.

 

[mrbill]

21 m/s is the velocity in the y direction. The resultant velocity vector (the velocity at which the car hits the ground) would be the sum of the vectors in the x and y directions.

The resultant vector would be: 12i -21j
The car would hit the ground at sqrt ( 21^2 + 12^2) m/s

The direction is given by arctan (21/12) S of E

 

[Drakon25th]

oohh, i see, thanks

 

[Drakon25th]

one question, could i use the formula:
sqrt ((.5Vo^2+gho-ghf)/.5) ?

what i mean:

sqrt ((.5 (20m/s)^2 + 9.8m/s^2 * 10m)/.5) = 24.4 m/s

Or would that not be right?