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 h = 1.5 m above the cars,and the horizontal distance he must clear is L = 17 m.

m/s

(b) If the ramp is now tilted upward, so that "takeoff angle" is 10° above the horizontal, what is the new minimum speed?

m/s

Answer 1

Height h = 1.5 mHorizontal distance L = 17 m(a). Time taken to reach the ground t = √[ 2h / g ]= 0.5532 sMinimum speed v = L / t= 30.72 m / s(b). In vertical direction :-----------------------Let the required speed be VInitila velocity u =V sin 10=0.1736 VAccleration a = -9.8 m /s ²Displacment S = h= 1.5 mFrom the relation S = ut + ( 1/ 2) at 21.5 = 0.1736 V t -4.9 t ² --------( 1)Also in horizonal direction , L = ( V cos 10 ) x tFrom this t = L / V cos 10= 17 / 0.9848V --------( 2)From eq( 1) and ( 2) ,1.5 = 0.1736 V [ 17 / 0.9848 V ] - 4.9 [ 17/0.9848 V ] ²1.5 = 2.996 - 1460.15 / V ²1460.15 / V ² = 2.996 - 1.5V ² = 1460.15 / ( 2.996-1.5)= 976.03V = 31.24 m / s