# Flying car

[car]
_o_o_____
.............|
.............| 1.48 meters high
________|_______________________(must land here)

.............|------21.1 meters--------|

A stunt driver wants to make his car jump over a distance of 21.1 meters below a horizontal ramp

With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is h = 1.48 m above the ground and the horizontal distance he must clear is d = 21.1 m.

What is the new minimum speed if the ramp is now tilted upward, so that "takeoff angle" is 11.6° above the horizontal, and nothing else has changed.

Can someone please give me some tips on how to start on these 2 problems?

Thank you very much for your help

Last edited:

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dextercioby
Homework Helper
Yes,u can start by applying the laws of kinematics of movement with (presumably) constant acceleration in free fall.

Post some ides,equations...

Daniel.

Here is what i have so far

Initial Distance Xo = 0 m
Final Distance X = 21.1 m
Acceleration = 0 m/s^2

Initial Height Yo = 1.48m
Final Height Y = 0 m
Acceleration = -9.8 m/s^2

I have tried applying the informaiton above to the position function and velocity functions however I end up with 2 unknown variables time and the initial velocity of distance and height

Distance

X = (1/2)(0)(t^2) + Vo(t) + 0
X = Vo(t)

Height

Y = (1/2)(-9.8)(t^2) + Vo(t) + 1.48

Y = (-9.8)t + Vo

Last edited:
dextercioby
Homework Helper
hotmail590 said:
X = Vo(t)

Y = (1/2)(-9.8)(t^2) + Vo(t) + 1.48

Y = (-9.8)t + Vo
Now come u got 3 equations,two of which for "Y"??The last one is completely wrong...

Daniel.

I have found my mistake. The initial velocity of height is 0m/s. If i just used the position function and plug all the values above in, then i will have found out the time for the car to land would be .5496 seconds. Then plug the time and the rest of the values for Distance, i would find the inital velocity of distance = 35.6985 m/s

dextercioby