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 22.8 mabove the river, while the opposite side is a mere 5.60 m above the river. The river itself is a raging torrent 65.0 m wide.

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?

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

Explanation / Answer

Vertical height difference between 2 edges = 22.8 - 5.60 = 17.2 m

So Car has to vertical 17.2 downwards with no initial vertical velocity as it was going horizontally.

Using H = uy*t - gt^2 //2

- 17.2 = 0 - 9.81 t^2 /2

t = 1.87 sec

So car have 1.87 sec to clear 65 m horizontally.

v = d/t = 65 / 1.87 = 34.71 m/s

Using work energy theorem,

mgh = mv^2 / 2 - mu^2 /2

9.81 x (22.8 - 5.6) = v^2 /2 - 34.71^2 /2

v = 39.27 m/s

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