Posted by **Somoan** on Friday, December 6, 2013 at 11:55pm.

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
a= 37m/s

b= 20m/s

- 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

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

Vyi-2.92m/s

Vinitial=23.94m/s

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