Finding the time for a splash to reach a bystander on a bridge

[garcia1]

Homework Statement

My problem is the last one, but it is a continuation of these first two problems, so I thought I'd include them.

In Mostar, Bosnia, the ultimate test of a
young man’s courage once was to jump off
a 400-year-old bridge (now destroyed) into
the River Neretva 27 m below the bridge.
How long did the jump last? The accelera-
tion of gravity is 9.8 m/s2 .
Answer in units of s.


How fast was the diver traveling upon impact
with the river?
Answer in units of m/s.

If the speed of sound in air is 343 m/s, how
long after the diver took off did a spectator
on the bridge hear the splash?
Answer in units of s.

Homework Equations

I used x = 1/2(Vo +V)T

The Attempt at a Solution

I found the values T = 2.3474 and V = -23.00 m/s for the 1st and 2nd portion respectively. On the third portion, I set the Vo to 0 m/s, and the distance as 27m as per the first problem. I also used V = 343 m/s from the last portion.

By plugging these values into equation x = 1/2(Vo + V)T, and solving for T, I got the answer .157.

Adding this to 2.3474s, which was the time it took for the diver to reach the water, I got the equation T = 2.3474s+.157s = 2.50s

This was wrong though, so I want to find out what I'm doing wrong here.

 

[gneill]

For a constant velocity, u, the distance traveled is d = u*t.

NOTE: It might be of interest to note that the Mostar Bridge has been rebuilt, mostly using the original material from the destroyed bridge. Young men can once again challenge their nerve and threaten their continued virility by jumping 27m into a cold river.

 

[garcia1]

I got the answer right, yet I'm interested to know why the acceleration vector is not included in the equation for constant velocity in this case. For doesn't the vector in this problem lie upon the y axis, which would be affected by the force of gravity?

 

[gneill]

Sound waves don't have weight.

 

[rwisz]

It seems like you are on the right track with your calculations. However, it is important to make sure that you are using the correct equations and units for each part of the problem.

For the first part, finding the time for the jump to last, you can use the equation d = (1/2)gt^2, where d is the distance (27m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Solving for t, we get t = √(2d/g) = 2.3474 seconds, as you calculated.

For the second part, finding the speed of the diver upon impact with the river, you can use the equation v = gt, where v is the final velocity, g is the acceleration due to gravity, and t is the time. Plugging in the values, we get v = (9.8 m/s^2)(2.3474s) = 23.00 m/s, as you calculated.

For the third part, finding the time it takes for a spectator on the bridge to hear the splash, we need to use the equation v = d/t, where v is the speed of sound (343 m/s), d is the distance (27m), and t is the time. Solving for t, we get t = d/v = 27m/343 m/s = 0.079s.

To find the total time for the splash to reach the bystander on the bridge, we can add the times calculated in the first and third parts, since they occur one after the other. So the total time would be 2.3474s + 0.079s = 2.4264s.

Make sure to always check your units and use the correct equations for each part of the problem. Good job on your attempt!