How Deep Is the Lake After Dropping a Ball from 30 Meters?

Thread starter miamiheat5
Start date Feb 19, 2007
Tags

Ball

AI Thread Summary

A heavy ball dropped from a height of 30 meters hits the water with a velocity of 24.3 m/s. After impacting the water, it continues to sink at this constant speed for 10 seconds. The calculation shows that the lake's depth is 243 meters, derived from multiplying the sinking velocity by the time taken to reach the bottom. The physics involved includes using the equations of motion to determine the final velocity and the distance traveled. Therefore, the lake is determined to be 243 meters deep.

Feb 19, 2007

miamiheat5

A heavy ball is dropped into a lake from a height of 30.0m above the water. it hits the water with a certain velocity and continues to sink to the bottom of the lake at this same constant velocity. It reaches the bottom of the lake 10.0s after it was dropped. How deep is the lake.

And the solution i got:

vf^2=vi^2+2ad
vf^2=2(-9.81)(30.0)
vf=24.3m/s

d=vt
=(24.3)(10)
=24.3m deep

Physics news on Phys.org

Feb 19, 2007

chaoseverlasting

24.3*10=243m

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...

View full post »

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »