A stunt driver wants to make his car jump over eight cars parked side by side be

Question

A stunt driver wants to make his car jump over eight cars parked side by side below a horizontal ramp.

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 1.5 degree above the horizontal, what is the new minimum speed? m . b-If the ramp is now tilted upward, so that takeoff angle is 10 m above the cars, and the horizontal distance he must clear is 20

Explanation / Answer

Well the key to this problrm is unnderstanding that the timeis the same for the horizontal and vertical motion. to start the problem off u have to imagine that the object isat free fall from rest and that would give you the time t such that y-y0=v0t +.5at2 1.5=0+4.9t2 t= .553283s then you can use the same t for the horizontal motion x-x0 =v0 t+.5at2 20=v0 (.553283)+0 v0=36.14 m/s ok, for the second part you can use this eqation x-x0=v02/g sin2 then the v0 is going to be 389 m/s Well the key to this problrm is unnderstanding that the timeis the same for the horizontal and vertical motion. to start the problem off u have to imagine that the object isat free fall from rest and that would give you the time t such that y-y0=v0t +.5at2 1.5=0+4.9t2 t= .553283s then you can use the same t for the horizontal motion x-x0 =v0 t+.5at2 20=v0 (.553283)+0 v0=36.14 m/s ok, for the second part you can use this eqation x-x0=v02/g sin2 then the v0 is going to be 389 m/s 20=v0 (.553283)+0 v0=36.14 m/s ok, for the second part you can use this eqation x-x0=v02/g sin2 then the v0 is going to be 389 m/s

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