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A ball is thrown upwards with an initial velocity of 16 m/s. What is the velocity of the ball 1.5 s later?
After the ball leaves the hand of the thrower, what is the value of acceleration acting on the ball as it is rising upwards?

9.8x1.5 = 14.7. 16-14.7=1.3 m/s after 1.5s.
The value of acceleration acting on the ball is -9.8 m/s, due to gravity.

Is this correct?

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Jim

  1. V = Vo + g*t = 16 - 9.8*1.5 = 1.3 m/s.
    Your answers are correct.

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    Henry

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