# How Do Two Divers' Speeds Compare When Hitting the Water from the Same Platform?

• darklich21
In summary, two divers jump from a 2.70 meter platform at different velocities and times. By using the equations Vf^2= Vi^2 + 2ad and Y=Y0 + ViT - 0.5gt, their final velocities and times to reach the water can be calculated. The diver who walks off the platform has a final velocity of 7.09 m/s and reaches the water in 0.74 seconds. The diver who jumps off the platform has a final velocity of 8.85 m/s and reaches the water in 0.92 seconds. Therefore, the diver who walked off the platform reaches the water first.
darklich21

## Homework Statement

Two divers jump from a 2.70meter platform. One jumps upward at 1.60meters/second and the 2nd steps off the platform as the first passes it on the way down. What are their speeds as they hit the water? Which hits the water first and by how much?

## Homework Equations

Vf^2= Vi^2 + 2ad (Vf being final velocity and Vi being initial velocity)
Y=Y0 + ViT - 0.5gt (Y and Ynaught being the distance in the y-axis direction, T being time, and g being gravity)

## The Attempt at a Solution

So i used the equation Vf^2=Vi^2 + 2ad to find the speed at which the diver that walked off the platform hit the water:

Vf^2= 0 + 2(-9.8m/s^2)(0-2.7m) = sqroot[(-18.6m/s^2)(-2.70m)]= 7.09m/s

I'm somewhat sure that to find the speed of the other diver I have to use that 2nd equation that I provided and 1 more additional equation, but yeah, I am stuck. Can anyone help? Equations and explanations are much appreciated, thanks

Oh, and off the bat I'm going to say the diver that jumped from the platform will reach the water first as opposed to the diver that walked off. But I don't know how to show that, so yeah...

Using the second equation, find the time taken by both of them.
For both of them, displacement is the same. Initial velocity for walker is zero.

Alright, so for the diver that jumps off, I used the equation Y=Y0 + Vit -0.5gt^2. Using some algebraic manipulation, t= [Vi +-sqroot(Vi^2 +2Y0g)]/g

1.6m/s +- sqroot((1.6^2)+2(2.7m)(9.8m/s^2))/9.8m/s^2. I get 2 answers, one of them is t=0.92s and the other is t=-0.59s, so I am rejecting the negative and accepting 7=0.92s as the time for the jumping diver.

For the diver that walks off, I used the same equation, except i made Vi=0. and I got 0.74s as my positive time.

So I'm guessing this proves the jumping diver gets to the water before the diver that walks off due to short time right?

And what about the speed for both of them hitting the water? Was my first equation right in doing that?

let me know if i did something wrong

For jumping diver t = 0.92 s to reach the watr.
For walking diver t = 0.74.
So who reaches the water first?
Using the first equation you can find their final velocities.

Based on the shorter time for the walking diver, he reaches the water first.

I got the final velocity of the step off diver correct, but the final velocity of the jumping diver was wrong, this is what I did.

Vf^2= 0 + 2(9.8m/s^2)(2.70m+1.6m)
Vf= 9.18m/s

I think my d in the equation is wrong, what do you think?

Initial velocity of the jumper is 1.6 m/s in the upward direction. When he crosses the starting point, his velocity is 1.6 m/s.
So vf = vi + g*t where t = 0.74 s

Alright so, Vf= Vi +gt
Vf= 1.6m/s + (9.8m/s^2)(0.74s)(0.74s)
Vf= 1.6 + 7.252
Vf= 8.852m/s

But if I'm dealing with the jumping diver, why am I using the time from the diver that's walking off? shouldn't I use t=0.92s?

That was a typo on my part, I didn't mean to type it twice. But yeah, why am I using the t=0.74s. Why am I not using t= 0.92s?

wait I am confused, how is it the same, I previously found 2 different times from which both swimmers will hit the water.

For walker
vf = 0 + g*t = 9.8* 0.7274
For jumper
vf = vi - g*t Here vi = 1.6 and t = 0.9233

## 1. What is motion in a straight line?

Motion in a straight line is when an object moves along a straight path, without any changes in direction.

## 2. What is the difference between speed and velocity?

Speed is the distance an object travels in a given amount of time, while velocity is the speed of an object in a specific direction.

## 3. How is acceleration calculated in motion in a straight line?

Acceleration in motion in a straight line is calculated by dividing the change in velocity by the change in time.

## 4. What is the difference between uniform motion and non-uniform motion?

Uniform motion is when an object moves at a constant speed, while non-uniform motion is when an object's speed changes over time.

## 5. How does Newton's First Law of Motion apply to motion in a straight line?

Newton's First Law of Motion states that an object will remain at rest or in uniform motion unless acted upon by an external force. This applies to motion in a straight line, as an object will continue to move in a straight line at a constant speed unless a force acts upon it.

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