A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 when the hand is 2.10 above the ground. How long is the ball in the air before it hits the ground?

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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 when the hand is 2.10 above the ground. How long is the ball in the air before it hits the ground?

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Ok, what's the equation of motion here? Let u = initial velocity s = initial position g = acceleration due to gravity. Then let y(t) be the position of the particle at time t, and v(t) its velocity at time t in the y direction. We have that \[ v(t) = u - gt \] and \[ y(t) = s + ut - \frac{1}{2}gt^2 \] Now use that and the data given your equation to find the time t for which y(t) = 0; because that is when it reaches the ground.
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