A stunt driver wants to make his car jump over 8 cars parked side by side below

Question

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 m.)

[B] If the ramp is now tilted upward, so that "takeoff angle" is 9.0? above the horizontal, what is the new minimum speed?

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp. With what minimum speed must he drive off the horizontal ramp? f the ramp is now tilted upward, so that "takeoff angle" is 9.0? above the horizontal, what is the new minimum speed?

Explanation / Answer

a). horizontal distance travelled = V*t => V*t = 22 vertical distance travelled = u*t + 0.5*g*t^2 => 1.5 = 0 + 0.5*9.8*t^2 => 1.5 = 4.9*(22/V)^2 => 0.306 = 484/V^2 => V = 39.77 m/s b). horizontal velocity = v*cos9 vertical velocity = v*sin9 horizontal distance travelled = (v*cos9)*t =>(v*cos9)*t = 22 also, vertical distance travelled = u*t + 0.5*g*t^2 1.5 = (-v*sin9)*t + 0.5*9.8*t^2 => 1.5 = (-v*sin9)*[22/(v*cos9)] + 4.9* [22/(v*cos9)]^2 => 1.5 = -3.484 + 2431.09/v^2 => v = 22.09 m/s

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