• anonymous
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?
  • Stacey Warren - Expert
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  • chestercat
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  • JamesJ
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.
  • anonymous
thank you

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