A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the bottom 15.4 cm of the jump?
I've been told that I need to use the height of the jump to solve for V. Then I need to use that V in the equation "x = Vt -(1/2)gt^2"
to solve for t.
But I'm confused about getting the V. Don't I need time in order to find V?
3 answers
Yes, you use the jump height to find the initial velocity of the jump. Then,you can use the equation you have shown with the the result, x, being set to 15.4cm. Solve for t. This will give the time in the bottom 15.4cm for the upward motion. The player will also spend the same amount of time on the downward part of the jump. Be sure to convert units as needed.
But don't I need time in order to find the initial velocity?
You can combine some of the constant acceleration equations to get one that does not have time:
(final velocity)^2 = (initial velocity)^2 + (2 * a * s)
where: a=acceleration, s=distance
(final velocity)^2 = (initial velocity)^2 + (2 * a * s)
where: a=acceleration, s=distance
3 answers