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.
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