Question
Acar comes to a brigde during a storm and finds the bridge washedout. The driver must get to the other side, so he decides totry leaping it with his car. The side of the road the car ison is 21.3m above the river, while the opposite side is amere1.8 m above the river itself is raging torrent 61.0 m wide. a) how fast should the car be traveling at the time it leavesthe road in order just to clear the river and leand safely on theopposite side? b) what is the speed of the car just before it lands on theother side? a) how fast should the car be traveling at the time it leavesthe road in order just to clear the river and leand safely on theopposite side? b) what is the speed of the car just before it lands on theother side?
Explanation / Answer
Edit: I used 1.3 meters instead of 1.8 meters as the lowerheight. Using y=v*t+1/2*a*t^2 (y representing vertical distance) y=21.3-1.8=19.5 meters a=9.8m/s^2 v(initial)= 0m/s [I'm assuming that the car is noton a ramp or grade as it approaches the bridge] Plugging into the above kinematic equation: 19.5=0+9.8/2*t^2 3.9796s^2=t^2 t=1.995ss That's the amount of time it takes for the car to dropthe 19.5 meters into the water (neglecting all dragforces). In order to determine the speed of the car, use the constantacceleration formula. D=t*(Vi+Vf)/2 (Again, you completely neglectdrag forces) We know D=61m and t=1.995s Since the car is not accelerating in this model, Vf=Vi. The equation simplifies to D=v*t or D/t=v So, 61m/1.995s=30.576 m/s As for part b, the horizontal velocity should remain thesame: 30.576 m/s forward The vertical velocity should be 9.8m/s^2*1.995s=19.551down Using vector addition (90 degree angles):(30.576^2+19.551^2)=36.292 m/s In case you need it, the angle of the velocity vector is -28.7degrees below the horizontal direction of motion. [Taken fromArctan(-19.551/30.576)= -32.596