Aperson is jumping off a bridge onto the top of a car that is passing underneath. suppose that the top of the bridge is h=10 meters above the car and the car is moving at a constant speed of v=30 mi/h the person wants to land in the middle of the car. how far from the bridge should the car be when the person jumps. express the equation in variables then numerically

18.96 m

Explanation:

Height from person jump, h = 10 m

Let it takes time t to reach to the car.

Use second equation of motion

Here, a  g = 10 m/s^2 , u = 0, h = 10 m  

By substituting the values, we get

10 = 0 + 0.5 x 10 x t^2

t = 1.414 s

The speed of car, v = 30 mi/h = 13.41 m/s

Distance traveled by the car in time t , d = v x t = 13.41 x 1.414 = 18.96 m

So, the distance of car from the bridge is 18.96 m as the man jumps.

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

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