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 18.4 m above the river, while the opposite side is only 2.3 m above the river. The river itself is a raging torrent 51.0 m wide. Part A 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. v0 = SubmitMy AnswersGive Up Part B What is the speed of the car just before it lands on the other side?

Explanation / Answer

Vertical motion is 0 whilst the car is in contact with the road. When it leaves the road, it is falling under gravity.

The equation of motion for this case is

s = ut + 0.5at^2

where, s is the distance

u, initial velocity

a, acceleration

t, time

s = 18.4-2.3 = 16.1 m

u = 0

a = 9.81 m/sec²

So you have

16.1 = 0 + 0.5*9.81*t^2, giving

t^2 = 16.1/(0.5*9.81)

t^2 = 3.28

t = sqrt(3.28) = 1.81 s

The car has 1.81 sec. to reach the other side. If its speed when it starts the jump is v m/sec, then it has to cover 51 m in that time.

51 = v(1.81)

v = 51/1.81 = 28.17 m/s

There is no acceleration, and you do not mention any resistance to the motion, so theoretically it should be travelling at the same speed until it touches down on the far side.

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