A car comes to a bridge during a storm and finds the bridge washed out. The driv

Question

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 23.0 m above the river, while the opposite side is a mere 2.1 m above the river. The river itself is a raging torrent 50.0 m wide. How fast should the car be traveling just as it leaves the cliff in order just 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

Given that,

vertical distance = y= 2.1 m ; y0 = 23 m ;

x = 50 m ;

Let vx be the speed with which the car levaes the cliff.

we know that, speed = dist/time, so vx x t = 50

t = x / vx

We know from third eqn of motion we know,

y = yo - 1/2 g t2

yo - y = 1/2 g t2 putting t = x /vx we get, and y =  yo - y = 23 - 2.1 = 20.9 m

y = 1/2 g x2 / vx2

vx = x sqrt (g / 2 y ) = 50 x sqrt ( 9.8 / 2 x 20.9) = 24.2 m/s

Hence, vx = 24.2 m/s

(b)The speed just before it lans on the other side will be given by:

v = sqrt ( x2 g / 2 y + 2 g y ) = sqrt ( 2500 x 9.8 / 2 x 20.9 + 2 x 9.8 x 20.9) = 31.55 m/s

Hence, v = 31.55 m/s

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