During a storm, a car traveling on a level horizontal road comes upon a bridge t

Question

During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 19.7 m above the river, while the opposite side is only 2.0 m above the river. The river itself is a raging torrent 62.0 mwide.

Find:

1. How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side?

Express your answer with the appropriate units.

2. What is the speed of the car just before it lands on the other side?

Express your answer with the appropriate units.

Explanation / Answer

Ignoring air resistance and wind speed.

the car drops ( 19.7 - 2.0 = ) 17.7 meters

with gravity (A) at 9.8 m/sec^2 that gives us a time of

distance = 1/2*g*t^2

t = sqrt ( 2 * 17.7 / 9.8 ) = sqrt (3.61) = 1.9 seconds

so 62 m in 1.9 seconds or ( (62/1.9) * (36000/1000) ) = 117.47 km/h

The speed of the car just before it lands

the horizontal speed is 117.47 km/h (ignoring wind resistance and all that)

the vertical speed is V = a*t

so V = 9.8*1.9 = 18.62 m/s or 67.03 km/h

Treat it as a vector problem Vcar = sqrt ( 117.47^2 + 67.03^2 ) = 135.25 km/h or 37.57 m/s

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