Vertical Jump Analysis: Calculating Time and Take-Off Speed

Click For Summary
SUMMARY

The discussion focuses on calculating the time a volleyball player is in the air and their vertical take-off speed after jumping to a height of 85 cm. The initial attempt used the kinematic equation d = v1(t) + 1/2(a)(t^2) but incorrectly assumed the initial velocity was zero. The correct approach involves recognizing that the time to ascend is equal to the time to descend, and the player’s speed at the peak height must be determined to solve for both time and take-off speed accurately. The final answer for the time in the air should be 0.83 seconds.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of free-body diagrams
  • Basic concepts of projectile motion
  • Familiarity with gravitational acceleration (9.81 m/s²)
NEXT STEPS
  • Learn how to derive vertical take-off speed using kinematic equations
  • Study the principles of projectile motion and time of flight
  • Explore the concept of initial velocity in free-fall problems
  • Practice solving similar problems involving vertical jumps and heights
USEFUL FOR

Students studying physics, coaches analyzing athletic performance, and anyone interested in the mechanics of vertical jumps in sports.

KaseyK
Messages
8
Reaction score
0

Homework Statement



A certain volleyball player can jump to a vertical height of 85 cm while spiking the volleyball.


Homework Equations



(a) How long is the player in the air for?
(b) What is the player's vertical take-off speed?

The Attempt at a Solution



Firstly I converted 85 cm to m, which is 0.85m.

I then drew a free-body diagram to help me visualize what was happening with the player. Sorry I can't draw it on here.

I then used the kinematic equation:

d=v1(t) + 1/2(a)(t^2)
Because I assumed that the inital velocity was zero (because he hasn't jumped up yet) the equation was left with:

d=1/2(a)(t^2)
.85x2 = 1/2(9.81)t^2
t = 0.61 seconds

The answer should be 0.83 seconds. Where did I go wrong?

I haven't attempted 'b' because I need to figure 'a' out first.

Thanks in advance for your help.
Kasey
 
Physics news on Phys.org
Initial velocity is not 0 because if she started on the ground, how did she get in the air? You have two variables, v0 and t.
 
Okay did I at least use the right kinematic equation?

Is there anything else you can hint to, to send me in the right direction?

Thanks
 
There are multiple ways of doing this. Yes your way is valid. Remember the time it takes to come down is the same as the time it takes to come up.
 
You can't calculate it like that because you don't know his initial speed yet. What you do know is his speed when he reaches 85cm height. What is it? How long does it take him to fall from 85cm back to the ground?

Edit: I made an assumption here that you have to solve it using kinematics. If this is not so then there is an easier way!
 
Last edited:
Because you have two unknowns, you need two equations. For the second equation, remember that the velocity has been reduced to zero at the top of the jump.
 
Cyosis said:
You can't calculate it like that because you don't know his initial speed yet. What you do know is his speed when he reaches 85cm height. What is it? How long does it take him to fall from 85cm back to the ground?

Edit: I made an assumption here that you have to solve it using kinematics. If this is not so then there is an easier way!

No I don't need to solve it using kinematics. I can use any way I want, I just can't figure either way out...ugh.
 
We can do it multiple ways, but let us first do it in the order you're asked to do it.

You need to answer this. What is the speed of the player at 85cm and how long does it take for him to fall 85cm after reaching that height?
 
Sorry when I said your way is valid, I was talking about your equation. Sorry for any confusion that it caused.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
893
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
908
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K