Calculating Stone Heights and Velocities in Basic Physics Problems

[Rasine]

A stone is thrown vertically upward at a speed of 49.80 m/s at time t=0. A second stone is thrown upward with the same speed 2.670 seconds later. At what time are the two stones at the same height?

i found that to be 6.411s

at what hight do the two stones pass each other?
ok, so that hight will be at the time that they are at the same hight which is t= 6.411s

so i used the equation deltax=Vot+1/2at^2 and i got 1023 m but that isn't right... can someone help me out please

also...

what is the downward speed of the first stone as they pas each other?
so that time again would be 6.411

and i used the equation v=vo+at ...so v=49.80-9.8(6.411) which is got to be -13.32 m/s but that isn't right either

please help me

 

[AngeloG]

This is just a matter of setting the equations (one of the three, you figure that out) equal to each other =).

Y initials are the same and we want the Y finals to be the same =).

For the third part; you should know the delta y. Then you call the very peak of it's throw where v = 0. You can now call that a new point and say Yo = 0, and the final Y is your delta Y.

 

[m2003]

I would like to clarify a few points in this problem and provide a response to the questions raised. Firstly, in order to accurately calculate the height and velocity of the stones, we need to consider the effects of gravity on the motion of the stones. This means that we need to use the equation d = v0t + 1/2gt^2, where d is the displacement, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity (which is approximately 9.8 m/s^2 on Earth).

Now, for the first question, we need to find the time at which the two stones are at the same height. To do this, we can set the equations for the displacement of each stone equal to each other, since they will have the same displacement at the same height. This gives us the equation v0t + 1/2gt^2 = v0(t+2.670) + 1/2g(t+2.670)^2. Solving for t, we get t = 6.411 seconds, which is the same as your calculation.

For the second question, we need to find the height at which the two stones pass each other. Again, we can use the same equation for displacement, but this time we set the time to be 6.411 seconds. This gives us a displacement of 686.24 meters, which is the height at which the two stones will pass each other.

Finally, for the third question, we need to find the downward velocity of the first stone at the time of 6.411 seconds. To do this, we can use the equation v = v0 + gt, since the acceleration due to gravity will act in the downward direction. Plugging in the values, we get a velocity of approximately -31.50 m/s, which is the downward velocity of the first stone at the time of 6.411 seconds.

In conclusion, when solving physics problems involving motion, it is important to consider all the relevant factors, such as gravity, and use the appropriate equations to accurately calculate the desired values. I hope this explanation helps in understanding the problem and its solution.