Anyone good at Physics? Need help with a problem

[ddollar]

So my friend has physics hw and for whatever reason she doesnt get this problem i dont tak physics so i have no clue any one here know the answer to this

A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.6s later, if the speed of sound is 340m/s, how high is the cliff?

 

[MalcolmJID]

1224m.



If the speed is 340m/s, you multiply by seconds (s) to get just m left. So 340*3.6= 1224m

 

[EricNau]

Perhaps I'm missing something, but I believe the answer is:

1,224 m

...Seems pretty straightforward to me.

 

[ddollar]

1224m.



If the speed is 340m/s, you multiply by seconds (s) to get just m left. So 340*3.6= 1224m

lol that is what i said but she said something about gravitional motion takes effect i have no clue, is she wrong?

 

[swiftaw]

you're both missing something.

That would be the height if the rock hit the water now, but first it actually has to reach the water.

Thus

Time to reach the water + time for sound to get back to you = 3.6 seconds

So, let g = gravitational acceleration (approx. 10m/s^2 but use whatever you learned it to be in class) and let x be the height of the cliff

Then, time to reach water = sqrt(2x/g) [using the fact that x = (1/2)gt^2]
and time for sound to get back to you is = x/340

Thus, solve sqrt(2x/g) + x/340 = 3.6 for x

 

[MalcolmJID]

you're both missing something.

That would be the height if the rock hit the water now, but first it actually has to reach the water.

Thus

Time to reach the water + time for sound to get back to you = 3.6 seconds

You raise a good point.

lol that is what i said but she said something about gravitional motion takes effect i have no clue, is she wrong?

If you're getting technical, then you'll need to use the suvat equations, of which, right now, I'm none too hot on (been too long out of practise). They will factor in the gravitational pull (i.e. accelerating from 0m/s at a rate of 9.81m/s^2 etc)

 

[ddollar]

You raise a good point.




If you're getting technical, then you'll need to use the suvat equations, of which, right now, I'm none too hot on (been too long out of practise). They will factor in the gravitational pull (i.e. accelerating from 0m/s at a rate of 9.81m/s^2 etc)

yes thats exactly what she said to me, how would she apply that equation in to finding the answer though?

 

[MalcolmJID]

yes thats exactly what she said to me, how would she apply that equation in to finding the answer though?

swiftaw has it in his post! Beat me too it. (I;m too drunk to be reliable right now )

 

[ddollar]

swiftaw has it in his post! Beat me too it. (I;m too drunk to be reliable right now )

so she would just plug 340 where 0 is in that equation then? Sorry guys i am not taking this class so i need to be really basic.

* she wants to know if this is the correct equation Y1=Vo T1 + 1/2 g(T1)^2

 

[swiftaw]

so she would just plug 340 where 0 is in that equation then? Sorry guys i am not taking this class so i need to be really basic.

* she wants to know if this is the correct equation Y1=Vo T1 + 1/2 g(T1)^2

Yes, which is the equation I used above. The initial velocity is zero, so it becomes Y1 = 1/2 * g *(T1)^2, or as I wrote it x = (1/2)*g*t^2.

Solving it for t says that the time it takes to hit the water if the cliff height is x is sqrt(2x/g)

 

[MalcolmJID]

so she would just plug 340 where 0 is in that equation then? Sorry guys i am not taking this class so i need to be really basic.

* she wants to know if this is the correct equation Y1=Vo T1 + 1/2 g(T1)^2

Yes correct equation. (I know that one as s=ut+(1/2a)t^2 where s=displacement, u = initial velocity, a=acceleration and t=time but because u=0 then it becomes s=1/2*a*t^2 and if you treat the height of the cliff as 'x', that becomes x=1/2*9.81*t^2. Rearrange for t, it becomes t=√(2x/a). Then for the time taken for the sound to travel back is x/340

Then solve for x from 3.6 = √(2x/9.81) + x/340

 

[swiftaw]

Yes correct equation. (I know that one as s=ut+(1/2a)t^2 where s=displacement, u = initial velocity, a=acceleration and t=time.)

Yep, that's what I learned at Winstanley 6th form.

 

[MalcolmJID]

Yep, that's what I learned at Winstanley 6th form.

As in, Winstantely 6th form near Wigan, U.K?

 

[swiftaw]

Then solve for x from 3.6 = t=√(2x/a) + x/340

Didn't I already write that?

As in, Winstantely 6th form near Wigan, U.K?

That's the one, born and raised in Ashton-in-Makerfield.

 

[MalcolmJID]

Didn't I already write that?



That's the one, born and raised in Ashton-in-Makerfield.

Yes you did already write that, I just wanted to get the old brain in gear and see if I could run through it in my head again!


I went to Byrchall High and then on to St. John Rigby 6th form myself. Live in Billinge! Small world.

 

[swiftaw]

Yes you did already write that, I just wanted to get the old brain in gear and see if I could run through it in my head again!


I went to Byrchall High and then on to St. John Rigby 6th form myself. Live in Billinge! Small world.

I went to Cansfield, Byrchall's old rival. Have some good friends who went to Byrchall though.

 

[MalcolmJID]

I went to Cansfield, Byrchall's old rival. Have some good friends who went to Byrchall though.

Still are rivals, them and St. Eddies! Byrchall has gone down the pan, as has much of Ashton (I deliver round there occasionally). Anyway, this is besides the point of the topic



ddollar, has your friend had any luck yet?

 

[EricNau]

you're both missing something.

That would be the height if the rock hit the water now, but first it actually has to reach the water.

Precisely why I was never good as word problems: I don't read them.

 

[ddollar]

Still are rivals, them and St. Eddies! Byrchall has gone down the pan, as has much of Ashton (I deliver round there occasionally). Anyway, this is besides the point of the topic



ddollar, has your friend had any luck yet?

she cant figure it out idk y she had all of those the formulas before i even asked u guys she knew what do but apparetnly she cant get the answer did any of you guys figure it out?

 

[swiftaw]

she cant figure it out idk y she had all of those the formulas before i even asked u guys she knew what do but apparetnly she cant get the answer did any of you guys figure it out?

The solution of sqrt(2x/g) + x/340 = 3.6 depends on what she needs to use for g

If use g = 10m/s^2 then x = 58.73m
If use g = 9.8m/s^2 then x = 57.66m
If use g = 9.81m/s^2 then x = 57.72m

 

[ddollar]

Another week another physics question lol sorry if these are annoying guys here is the question

A baseball is hit with a speed of 30.0mph at an angle of 49.0 degrees. It lands on the flat roof of a 11.0 ft-tall nearby building.
If the ball was hit when it was 1.5ft above the ground, what horizontal distance does it travel before it lands on the building?

 

[djellison]

No offense - but these are fairly simply mechanics problems. If she can't figure them out, she needs to get help from the course staff.

I'll give you a basic start - but no one should really be helping beyond that. Having the thing given to her on a plate, again, isn't going to teach her anything.

So...

Resolve the speed of the hit and the angle into vertical and horizontal velocity components (draw a triangle to help visualise this and figure out the basic trig using SOHCAHTOA ) Use basic equations to calculate it's time of flight using the vertical element to fly up from the height it was hit, slow to a stop, then fall back down to the roof. Then, take this time and multiply by the horizontal component to calculate how far it has traveled.