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A tennis player throws a ball vertically upwards at 4.90 m/s. How long, in seconds, does the ball take to return to his hand (assuming it has not moved since releasing the ball)?

Assume a gravitational acceleration of g = 9.8 m/s^2 and give your answer to 2 decimal places.

Respuesta :

If ball return to the hand of player then we can say that displacement of ball will be zero

so here we will have

[tex]\Delta y = v t + \frac{1}{2} at^2[/tex]

so here we will have

[tex]0 = 4.90(t ) + \frac{1}{2}(-9.8)t^2[/tex]

[tex]0 = 4.90- 4.90 t[/tex]

[tex]t = 1.0 s[/tex]

so it will take 1 second to come back in its hand

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