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 horiontal ramp?The vertical height of the ramp is 1.5m above the cars and the horizontal distance must be clear is 20m.(b)If the ramp now tilted upwards,so that "take off angle" is 7.0 above the horizontal,what is the new minimum speed?

Answer 1

tanA = 1.5/20 = 0.075,
A = 4.3 Deg.

a. Vf^2 = Vo^2 + 2gd,
Vo^2 = Vf^2 - 2gd,
Vo^2 = 0 - 2*(-9.8)*1.5 = 29.7,
Vo(v) = 5.45m/s. = ver. component.
Vo = 5.45 / sin4.3 = 72.7m/s = Inital
velocity.

Answer 2

wat about b?