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 h1 meters above the river, while the opposite side is lower and only h2 meters above the river. The river itself is a raging torrent x meters wide. 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? (enter an expression)

Explanation / Answer

let the initial velocity be "v" along the horizontal direction

consider the motion along the vertical direction

Y = vertical displacement = final position - initial position = h2 - h1

Voy = initial velocity along Y-direction = 0 m/s

t = time taken

a = acceleration = - g

using the equation

Y = Voy t + (0.5) a t2

h2 - h1 = 0 t + (0.5) (- g) t2

t = sqrt(2 (h1 - h2)/g) eq-1

consider the motion along the horizontal direction

Vox = initial velocity = v

x = width of river

t = time

using the equation

x = Vox t

using eq-1

x = v (sqrt(2 (h1 - h2)/g))

v = x sqrt(g/(2 (h1 - h2)))

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