Constant acceleration problem

  • Thread starter apiwowar
  • Start date
  • #1
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A basketball player grabbing a rebound jumps 76.1 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 14.4 cm of this jump and (b) in the bottom 14.4 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

so the only thing i thought of doing was finding the initial velocity

Vi^2 = -2ax
Vi= 3.86 m/s

now i have no idea what do to.
any help?
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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You need to find the time the jumper reaches height 76.1-14.4.
So you need a distance as a function of time formula.
 
  • #3
Up an down times are equal so begin by solving for descent data.
At 76.1cm, vertical velocity is zero.
Solve your second equation for s. Becomes your time equation.

Calculate velocity after falling 14.4cm. This can be plugged into vi in the time equation to determine seconds falling.

Begin at 0. Solve for -14.4cm. Apply to your time equation. Time 1 - time 0 = 'hang time'/2.

Do the same for -(76.1cm-14.4cm) to find time at 14.4 cm above the gym floor. Do the same for -76.1cm. Subtract again.

Make sure you reconcile up plus down times (multiply by two).
 

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