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 23.6 m above the river, while the opposite side is a mere 6.40 m above the river. The river itself is a raging torrent 66.0 m wide.

Part 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? vmin = ____________m/s

Part B What is the speed of the car just before it lands safely on the other side? v=____________m/s

Explanation / Answer

23.6m - 6.4m = 17.2m = difference in elevation.
assuming the 23.6m brige approach is horizontal,

d = at^2 / 2

time to free fall 17.2m = (2×17.2m÷9.81m/s²) = 1.872seconds

To cover 66m in 1.872s requires velocity

d = vt

66m/1.872s = 35.245 m/s

Part B

The horizontal velocity on landing is equal to launch velocity.

Velocity of car when it lands is the combined horizontal velocity plus the velocity of its accelerated drop:

v(y) = at
v(y) = 9.8 m/s^2 * 1.872s = 18.3456m/s

The total velocity is [(35.245m/s)²+(18.3456m/s)²] = 39.733m/s

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