What is the height of the cliff?

  • Thread starter Thread starter Nghi
  • Start date Start date
  • Tags Tags
    Ball
AI Thread Summary
A stone is dropped from a cliff, and after 2.18 seconds, a second stone is thrown downward with an initial speed of 31 m/s, both hitting the ground simultaneously. The first stone's velocity after 2.18 seconds is calculated to be 21.3858 m/s, and its distance traveled is found to be approximately 23.31 meters. The discussion highlights confusion about how to proceed with the second stone's motion and the relationship between their times of flight. Ultimately, the realization occurs that the second stone's time of flight is 2.18 seconds less than that of the first stone, which is crucial for solving the problem. The focus should remain on determining the height of the cliff using the correct equations for both stones.
Nghi
Messages
16
Reaction score
0

Homework Statement



A stone is dropped from a cliff; 2.18 s later another stone is thrown downward with an initial speed of 31 m/s. They reach the ground simultaneously. Find the height of the cliff.

Homework Equations



vf^2 = vo^2 + 2ax
x = vot + 0.5at^2
vf = vo + at

The Attempt at a Solution



i've been working on this for the past two hours, and it's driving me nutsss.

i found the speed of the stone in 2.18 seconds by using the equation vf = vo + at. (t = 2.18 seconds; a = 9.81 m/s^2; vo = 0) i found the velocity to be 21.3858 m/s.

with that information, i found how far the stone traveled with the equation vf^2 = vo^2 + 2ax. (vf^2 = (21.3858)^2; vo^2 = 0; a = 9.81) the answer: 23.310522 m.

after that, i don't know what else to do with stone b. I've been playing around with the first equation, since the two stones hit the ground at the same time. is it wrong to assume that the final velocities of the two would be the same?

i would appreciate it very much if someone could throw me a bone on what to do with the second stone because I'm at a loss.

thanks!
 
Physics news on Phys.org
You have the right equations, and are plugging in values correctly. But you seem to be calculating unnececessary things.

Is it in any way useful to calculate how far the first stone has traveled after 2.18s?

Focus on thinking about what you really want, i.e. the height of the cliff. Think of x as being the height of the cliff. Then plug in values for your equations for the first stone, and then the second stone.

Given both sets of equations, solve for x.
 
i'm still stuck on what i need to do... i don't know where to go after finding the speed of the first stone at 2.18 seconds. i know a = 9.81 m/s^2, speed of the first stone is 21.3858 m/s when the second stone is thrown with an initial velocity of 31 m/s. after that, everything is coming up blank. I've tried to set equations equal to each other, but they always cancel out the wanted variable.
 
AAAHHH I GOT IT I GOT IT! i forgot that the time for the stone b is 2.18 seconds less than stone a! yayyy!
 
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...
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...
Back
Top