A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.
A)With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 22..(I got 40 m/s)
B)If the ramp is now tilted upward, so that "takeoff angle" is 7.0∘ above the horizontal, what is the new minimum speed? (I cant figure out)
Physics - not ure bae, Sunday, May 3, 2015 at 4:58pm
Physics - Cole, Sunday, January 24, 2016 at 3:16pm
Okay, I was having problems with this and tried twisted the numbers around, not sure if right but it could be.
A) Vi=39.8 m/s
to make T=x/Vxi
This way you use substitution
When you look at the vectors you can relate the angle and the two velocity vectors using tan(7)=(Vyi/Vxi)
Substitute that back in and you get
Y=final Y so Y=-1.5m
X=final X so 22m
Using all of this you can find Vxi and from there get the other component as well as the answer
I find that the answer is...