# A ball is thrown up into the air with the initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed for the ball to reach its max height.

1 answer

###### Question:

A ball is thrown up into the air with the initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed for the ball to reach its max height.

## Answers

Answer:(a) h = 16.53 m (b) t = 1.83 sExplanation:Given that,The initial velocity of a ball, u = 18 m/sWhen it reaches to the maximum height, its final velocity v will be 0. Let it goes to a maximum height of h meters.Finding t using first equation of motion as follows :v = u +atHere, a = -g and v = 0[tex]t=\dfrac{u}{g}\\\\t=\dfrac{18}{9.8}\\\\t=1.83\ s[/tex]The time needed for the ball to reach its max height is 1.83 s.Let h is the maximum height. Using second equation of motion to find it :[tex]h=ut-\dfrac{1}{2}gt^2\\\\h=18(1.83)-\dfrac{1}{2}\times 9.8\times (1.83)^2\\\\h=16.53\ m[/tex]So, it will go to a maximum height of 16.53 m.

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