Question

A stunt driver wants to make his car jump over eight 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 1.5m above the cars, and the horizontal distance he must clear is 20m . wich is 36m/s but...

b)If the ramp is now tilted upward, so that "takeoff angle" is 14° above the horizontal, what is the new minimum speed???? how do I do this?

Answer 1

how long to fall 1.5 meters?
1.5 = .5 (9.81) t^2

how far horizontal ? 20 meters
so
u t = 20 where t = t from the vertical problem above
u = 20/that t

now do your vertical problem with vi = s sin 14 and u = s cos 14

h = vi t - 4.9 t^2 = 1.5
1.5 = (s sin 14)t - 4.9 t^2

20 = (s cos 14) t
so t = 20/s cos 14

1.5 = sin 14*20/ cos 14 -4.9(400/s^2 cos^2 14)
solve for s

Answer 2

how do i solve for s? I don't undestand!!!

Answer 3

1.5 = 4.99 - 2000/s^2
-3.49 = -2000/s^2
s^2 = 573
s = 24 m/s