Physics
posted by Somoan .
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)

a= 37m/s
b= 20m/s 
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
B)Y=Yi+VyiT+(1/2)a(t^2)
use X=Xo+VxiT
to make T=x/Vxi
This way you use substitution
Y=Yi+Vyi(X/Vxi)+(1/2)a(X/Vxi)^2
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=Yi+tan(7)X+(1/2)a(X/Vxi)^2
Y=final Y so Y=1.5m
X=final X so 22m
a=9.8m/s^2
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...
Vxi=23.76m/s
Vyi2.92m/s
Vinitial=23.94m/s
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