How fast does a diver hit the water when jumping off a 30m platform?

Click For Summary

Homework Help Overview

The problem involves a diver stepping off a 30m platform and seeks to determine the speed at which the diver hits the water. The subject area pertains to kinematics and the equations of motion under constant acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations related to constant acceleration, including attempts to recall and apply them correctly. There is an exploration of the relationship between time, distance, and final velocity.

Discussion Status

Some participants have provided guidance on the appropriate equations to use, while others have pointed out discrepancies in the original poster's calculations. There is an ongoing examination of the steps taken and the correctness of the values used.

Contextual Notes

There is mention of constraints regarding the memorization of equations as per the original poster's teacher's instructions. Additionally, some confusion arises from the mixing of time and speed in the responses.

Medgirl314
Messages
568
Reaction score
2

Homework Statement



A diver steps off a platform 30m above the water. With what speed does he hit the water?


Homework Equations


v=intial v+at

The Attempt at a Solution



I reached an answer of 2.47 seconds, but I can't see how I got there anymore! Is this answer correct?

Thank you very much!
 
Physics news on Phys.org
Hi Medgirl314! :smile:

You need to learn the standard constant acceleration equations.

In this case, you know vi and a and s, and you want vf

so the equation to use is … ? :wink:
 
Thanks for the reply!

I think I used several equations, but they faded a bit. Blasted erasable pens. Since it's a fairly new concept, my physics teacher doesn't want me to have them memorized yet. I think I used y=y0+v0+1/2 at^2 and t^2=2y/a .

The answer seems plausible. Did I get it right?

Thanks again, and thanks for the link!
 
(try using the X2 and X2 buttons just

Hi Medgirl314! :smile:

(try using the X2 and X2 buttons just above the Reply box :wink:)
Medgirl314 said:
… my physics teacher doesn't want me to have them memorized yet.

rubbish!

there's only 3 basic equations, memorise them and everything will be much simpler!
I think I used y=y0+v0+1/2 at^2 and t^2=2y/a .

you mean y=y0+v0t + 1/2 at2

(and the second one is just another version of the first one, with zero constants)
The answer seems plausible. Did I get it right?

(the question asked for a speed, and your answer is a time :confused:)

on this forum, you need to show your calculations, not just the answer :wink:
 
Oh! I'm sorry! My answer was actually 24.2 m/s. I gave you a different part that I used to find that answer. I think I used t=√2y/a, and got t=√2.3 m/9.8 m/s^2, giving a time of 2.47 seconds. Then I used v=vo+at , plugging in my numbers to get v=0+9.8 m/s^2(2.42 s), and then got my answer of v=24.2 m/s. Is this better? Thank you!
 
HI Medgirl314! :smile:

(never use . for multiplication … use x or * or brackets :wink:)
Medgirl314 said:
… I used t=√2y/a, and got t=√2.3 m/9.8 m/s^2, giving a time of 2.47 seconds. Then I used v=vo+at , plugging in my numbers to get v=0+9.8 m/s^2(2.42 s), and then got my answer of v=24.2 m/s. Is this better? Thank you!

yes, that's correct (you meant 30 not 3)

but quicker would have been to use vf2 = vi2 + 2as :wink:
 
Thank you! I must have used a dot for multiplication on my paper and didn't notice that it was 30, not 3.0!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
4K
Replies
7
Views
4K
Replies
6
Views
2K
Replies
6
Views
2K
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
9K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
8
Views
4K
Replies
6
Views
2K