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

Question

A 10000 car comes to a bridge during a storm and finds the bridge washed out. The 650 driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 24.0 above the river, while the opposite side is a mere 5.40 above the river. The river itself is a raging torrent 66.0 wide.

A) How fast 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 just before it lands safely on the other side?

Explanation / Answer

the object (car + man) has to travel 75.1 m horizontally and drop 31.2 - 2.29 = 28.91 m vertically to make it successfully. The total weight of the object is 10,000 + 650 = 10,650 N. which does not make any difference in this problem because, neglecting air resistance, all objects regardless of weight fall the same distance in the same time. x = 75.1 m. = v t, where x = horizontal distance the cars travels, v = speed of car as it leaves the cliff, and t = time in seconds the car is in the air y = 2.29 m. = 31.2 m. - 0.5 g t^2 where y = 2.29 m. = height of road on the opposite side of the river; 31.2 m. is the height of cliff that car leaves the road; g = 9.8 m/s^2 = acceleration due to gravity of the falling car; and t = time in seconds the car is in the air Solving the second equation: 2.29 = 31.2 - 0.5 g t^2 0.5 (9.8) t^2 = 31.2 - 2.29 t^2 = 28.91 / 4.9 = 5.9 t = SQRT(5.9) = 2.43 s. Solving the first equation: x = 75.1 m. = v t where t = 2.43 s. 75.1 = v (2.43) v = 30.9 m/s

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