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A track meet is held on a planet in a distant solar system. A shot-putter releas

ID: 1443798 • Letter: A

Question

A track meet is held on a planet in a distant solar system. A shot-putter releases a shot at a point 2.0 m above ground level. A stroboscopic plot of the position of the shot is shown in the figure, where the readings are 0.60 s apart and the shot is released at time t = 0. (a) What is the initial velocity of the shot in unit-vector notation? (b) What is the magnitude of the free-fall acceleration on the planet? (c) How long after it is released does the shot reach the ground? (d) If an identical throw of the shot is made on the surface of Earth, how long after it is released does it reach the ground?

Explanation / Answer

Here,   Vx * (0.60) = 5

=> Vx = 8.33 m/sec

Also, 4 = Vy * 0.60 - 1/2 * a * (0.60)2

             6 = Vy * 1.2 - 1/2 * a * (1.2)2

Solving these equations

=>    a)   initial velocity of shot =   8.33 i   + 8.33 j   m/sec

        b)     magnitude of free fall acceleration = 5.55 m/sec2

         c)      Here,   - 2 = 8.33t - 0.5 * 5.55 * t2

            =>      2.775t2 - 8.33t -2 = 0

          =>   t =   3.225 sec         -------------> time after it is released does the shot reach the ground

d)     Here,    - 2 = 8.33t - 0.5 * 9.81 * t2

           =>    4.905t2 - 8.33t -2 = 0

    =>     t = 1.9115 sec             ------------------>   on the surface of Earth time to reach ground

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