Comparing Two Stones: Solving the Mystery of Height and Time

AI Thread Summary
Two stones are thrown vertically upward at the same speed of 43.00 m/s, with the second stone launched 2.250 seconds after the first. To find when they are at the same height, four equations are established based on their heights and times. The first stone reaches a maximum height of 94.34 meters, and the time it takes to return to its starting point is calculated as 8.78 seconds. The time when both stones are at the same height is determined to be 5.5 seconds. The discussion continues with efforts to calculate the specific height at which they meet.
ZoomZoom6
Messages
2
Reaction score
0
I've been kicking myself in the head all day trying to figure this one out. I'm completely bewildered.:rolleyes: Please help meeee!


A stone is thrown vertically upward at a speed of 43.00 m/s at time t=0. A second stone is thrown upward with the same speed 2.250 seconds later. At what time are the two stones at the same height?

AND:

At what height do the two stones pass each other?

*Just need some help setting it up. I've never done an equation comparing two objects before.*
 
Last edited:
Physics news on Phys.org
ZoomZoom6 said:
*Just need some help setting it up. I've never done an equation comparing two objects before.*

For one object you can write an equation relating its height h1, and the time elapsed from the time it was thrown t1. You can write the same sort of equation relating h2 and t2. From here you have two equations and four unknowns (h1,t1,h2,t2)

A second stone is thrown upward with the same speed 2.250 seconds later means that t2=t1+2.25s. This is your third equation.

Think about the mathematical relationship between h1 and h2 when the two stones are at the same height. This should give you your fourth equation.

You will now have 4 equations and 4 unknowns, which means you can have enough equations to deduce all the unknowns.
 
atyy said:
For one object you can write an equation relating its height h1, and the time elapsed from the time it was thrown t1. You can write the same sort of equation relating h2 and t2. From here you have two equations and four unknowns (h1,t1,h2,t2)

A second stone is thrown upward with the same speed 2.250 seconds later means that t2=t1+2.25s. This is your third equation.

Think about the mathematical relationship between h1 and h2 when the two stones are at the same height. This should give you your fourth equation.

You will now have 4 equations and 4 unknowns, which means you can have enough equations to deduce all the unknowns.



I'm not sure I'm totally following you. Both H1 and H2 will be the same since that corresponds to the maximum height they reach, which I got to be 94.34 meters. I also calculated the time it took the first stone thrown to go up and back down to where it started from. t1=0 (starting point) & t1=8.78 seconds (finish point).


EDIT: I just solved at what time the two stones are at the same height, which is 5.5 seconds, by drawing a picture and working from there. I'm still having trouble trying to get the height where they're both at the same height.
 
Last edited:
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