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 650 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 24.0 m above the river, while the opposite side is a mere 5.00 m above the river. The river itself is a raging torrent 60.0 m wide.

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

Explanation / Answer

Motion along vertical or y-direction :

Voy = initial velocity = 0 m/s

a = 9.8 m/s2

Y = displacement = 24 - 5 = 19 m

t = time taken

using the equation

Y = voy t + (0.5) a t2

19 = 0 (t) + (0.5) (9.8) t2

t = 1.97 sec

Motion along X-direction ::

X = distance travelled = 60 m

t = 1.97 sec

V = speed of the car as it leaves = X/t = 60 / 1.97 = 30.5 m/s

Vf = Velocity just before the car lands

h = height above the other side = 19 m

m = mass = weight / g = 10000 / 9.8 = 1020.41 kg

using conservation of energy

Kinetic energy on bottom = Kinetic energy at top + potential energy

(0.5) m vf2 = (0.5) m v2 + mgh

vf2 = v2 + 2gh

vf2 = 30.52 + 2 (9.8) (19)

vf = 36.1 m/s

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