A 10000 N car comes to a bridge during a storm and finds the bridge washed out.

Question

A 10000 N car comes to a bridge during a storm and finds the bridge washed out. The 750 N driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.2 m above the river, while the opposite side is a mere 3.60 m above the river. The river itself is a raging torrent 64.0 m wide.

A. How fast in m/s should the car be traveling just as it leaves the cliff in order to clear the river and land safely on the opposite side?

B. What is the speed of the car in m/s just before it lands safely on the other side?

Explanation / Answer

a) i dont' think u need the weight of the car nor the driver at all just this y=0.5at^2, a is 9.8, and y=17.6, which is the actual height that the car travels, so solve for t, u get 1.9 second. then solve for vo in this x=vot, this is the horizontal movement equation, unliek the previosu one, which is vertical movement. So u get v=33.68m/s. b) use this equation 17.6=vf(1.9)-0.5a(1.9^2) a is still 9.8 so solve for vf, u get 18.57 m/s. This is teh speed that goes downward just before it lands on the other side. So u use pythragram r^2=x^2+y^2 to get teh final speed which is 33.46=square root (33.68^2+18.37^2), so vf=38.46 m/s To find the angel, or direction, use arc tangent (18.37/33.68) which si 28 degrees. so u get Right 28 degrees Down. Or in polar coordinate, its 332 degrees.

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