Motion Problem

[heartshapes]

I have a motion problem where you are dropping a rock of a bridge and you hear it hit the ground 5 seconds later.
I determined the height to be 400ft knowing that a(t)=-32ft/sec[sup:3amvl6aj]2[/sup:3amvl6aj] and that s(5)=0.

Then it says that you need to take into account the fact that sound does not travel instantaneously and to assume that sound travels at 1000 ft/s. Then you have to determine the height of the bridge if you hear it fall 5 seconds later.
I have no idea where to start. All of can think of is that the speed of sound has something to do with velocity?

Thank you.

 

[skeeter]

let t[sub:1cwrbh9k]1[/sub:1cwrbh9k] = time for the rock to drop
t[sub:1cwrbh9k]2[/sub:1cwrbh9k] = time for the sound to reach your ear
h = height of the drop

speed of sound at sea level at is approx 1100 ft/sec

$\displaystyle h = 16t_1^2$

$\displaystyle h = 1100t_2$

set the two expressions for h equal to each other ...

$\displaystyle 16t_1^2 = 1100t_2$

total time is 5 seconds ...

$\displaystyle t_1 + t_2 = 5$

solve the system for t[sub:1cwrbh9k]1[/sub:1cwrbh9k] or t[sub:1cwrbh9k]2[/sub:1cwrbh9k], then determine h.

you should get
$\displaystyle h \approx 350.6 \, ft$

 

[mmm4444bot]

… Then it says … to assume that sound travels at 1000 ft/s …

I worked this out while skeeter was posting, but I followed the instruction that you posted.

Check your typing. If the value 1000 is not a typographical error, then use it.

(I got 346.5 feet.)

 

[heartshapes]

Thank you both but I am still not getting the same answer that skeeter got. Also thank you it was suppose to be 1100 m/s. I should pay more attention.

I am ending up with 346.7 ft.

I did it so that 16(5-t[sub:3h0p6fcn]2[/sub:3h0p6fcn])[sup:3h0p6fcn]2[/sup:3h0p6fcn]=1100t[sub:3h0p6fcn]2[/sub:3h0p6fcn]
400-1132t[sub:3h0p6fcn]2[/sub:3h0p6fcn]+$\displaystyle 16t_2^2$=0

then I did the quadratic formula and got ... t[sub:3h0p6fcn]2[/sub:3h0p6fcn]= 70.3948 and 0.35513956 I then tried both as values of t[sub:3h0p6fcn]2[/sub:3h0p6fcn] and found that .355 was better. Then I plugged that in for t[sub:3h0p6fcn]2[/sub:3h0p6fcn] and got t=4.645 and found the height to be 346.7 but my two heights aren't equal when I plug into the original equations..

 

[mmm4444bot]

400 - 1132t[sub:1mebynse]2[/sub:1mebynse] + $\displaystyle 16t_2^2$=0

Arithmetic error somewhere while finding first-degree coefficient.

 

[Deleted member 4993]

Thank you both but I am still not getting the same answer that skeeter got. Also thank you it was suppose to be 1100 m/s. I should pay more attention.

I am ending up with 346.7 ft.

I did it so that 16(5-t[sub:3sb7izrl]2[/sub:3sb7izrl])[sup:3sb7izrl]2[/sup:3sb7izrl]=1100t[sub:3sb7izrl]2[/sub:3sb7izrl]

are your units consistent?

400-1132t[sub:3sb7izrl]2[/sub:3sb7izrl]+$\displaystyle 16t_2^2$=0

then I did the quadratic formula and got ... t[sub:3sb7izrl]2[/sub:3sb7izrl]= 70.3948 and 0.35513956 I then tried both as values of t[sub:3sb7izrl]2[/sub:3sb7izrl] and found that .355 was better. Then I plugged that in for t[sub:3sb7izrl]2[/sub:3sb7izrl] and got t=4.645 and found the height to be 346.7 but my two heights aren't equal when I plug into the original equations..

 

[skeeter]

$\displaystyle 16t_1^2 = 1100(5 - t_1)$

$\displaystyle 16t_1^2 = 5500 - 1100t_1$

$\displaystyle 16t_1^2 + 1100t_1 - 5500 = 0$

two roots using the quadratic formula ...
$\displaystyle t_1 \approx 4.68$
$\displaystyle t_1 \approx -73.43$

$\displaystyle t_1 \approx 4.68 \, sec$

$\displaystyle 16t^2 = h = 350.6 \, ft$

$\displaystyle t_2 = 5-t_1 \approx 0.32 \, sec$

$\displaystyle 1100t_2 = 350.6 \, ft$