Basketball hangtime physics problem

Click For Summary
SUMMARY

The discussion focuses on calculating the total time a basketball player spends in the top and bottom 13.3 cm of a 75.4 cm vertical jump. The correct approach for part (a) involved using the equation vf² = vi² - 2ay to determine the initial velocity (vi) as 1.6 m/s, leading to a total time of 0.3325 seconds in the top segment. For part (b), the incorrect application of dy = vit - 0.5gt² resulted in an erroneous time calculation. The conversation highlights the importance of considering both ascent and descent to accurately determine the total time spent at specific jump heights.

PREREQUISITES
  • Understanding of kinematic equations, specifically vf² = vi² - 2ay
  • Familiarity with the concepts of vertical motion and gravitational acceleration (g)
  • Knowledge of basic algebra for manipulating equations
  • Ability to interpret physics problems related to projectile motion
NEXT STEPS
  • Study the derivation and application of kinematic equations in vertical motion
  • Explore the concept of gravitational acceleration (g) and its effects on jump dynamics
  • Learn about the physics of projectile motion and its relevance to sports
  • Practice solving similar physics problems involving vertical jumps and time calculations
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, athletes interested in understanding jump mechanics, and educators teaching motion concepts in sports science.

wbetting
Messages
24
Reaction score
0

Homework Statement


A basketball player grabbing a rebound jumps 75.4 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 13.3 cm of this jump and (b) in the bottom 13.3 cm? Do your results explain why such players seem to hang in the air at the top of a jump?


Homework Equations



vf^2=vi^2-2ay
d=v/t

The Attempt at a Solution


i got part a correct by using vf^2=vi^2-2ay to find 1.6 m/s then i used d=v/t which i rewrote as t=2y/vi+vf and got .3325 for my answer which was correct. then part b came and i tried to use dy=vit-.5gt^2 and i got .0767 which was wrong
 
Physics news on Phys.org
Hi wbetting!

wbetting said:

The Attempt at a Solution


i got part a correct by using vf^2=vi^2-2ay to find 1.6 m/s then i used d=v/t which i rewrote as t=2y/vi+vf and got .3325 for my answer which was correct. then part b came and i tried to use dy=vit-.5gt^2 and i got .0767 which was wrong

You need to find the total time the basketball player is in both ascent and descent. Your equation will only give you one of those. :wink:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
25
Views
6K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 7 ·
Replies
7
Views
3K