A ball is thrown vertically into the air with an initial velocity of 19 m/s. What height will it travel upwards before it starts to reverse its direction?

Given parameters are: Initial velocity of the ball (u) =19m/s. The velocity of the ball at maximum height (v) = 0. g = 9.8m/s^2 Let us consider the maximum height as H. Consider the formula, 2gH = v^2– u^2 2 × (- 9.8) × H = 0 – (19)^2 – 19.6 H = – 361 H = 18.418 m. Therefore, The ball will travel a maximum height of 18.418m upwards before it starts to reverse its direction. Feel free to ping me for any further clarifications. Thanks.

Karan Mahendra Jain

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Physics Tutor • GCSE

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