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A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time did the skateboarder spend in the air and how far will be his final position from the bridge?

Sagot :

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

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