At what time do the two stones reach the same height?

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Two stones are thrown vertically upward at the same speed of 25.30 m/s, with the second stone released 1.860 seconds after the first. To determine when they reach the same height, the equations of motion for both stones must be set equal, taking into account the time difference. The first stone's height can be calculated using the equation X - Xo = Vox*t - 0.5*g*t^2, while the second stone's height must include the delay. Understanding the time to reach maximum height is crucial for accurate calculations. The discussion emphasizes the importance of correctly setting up the equations to find the point at which both stones are at the same height.
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A stone is thrown vertically upward at a speed of 25.30 m/s at time t=0. A second stone is thrown upward with the same speed 1.860 seconds later. At what time are the two stones at the same height?

what are the equations needed, and any help is greatly appreciated thanks
 
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Hi blueskadoo,

blueskadoo42 said:
A stone is thrown vertically upward at a speed of 25.30 m/s at time t=0. A second stone is thrown upward with the same speed 1.860 seconds later. At what time are the two stones at the same height?

what are the equations needed, and any help is greatly appreciated thanks

What have you tried so far?
 
i have tried setting them equal.
also, tried using the equation X-Xo= Vox*t + .5Ax*t^2
and Vx^2=Vox^2 + 2 Ax (x-xo)

i can't seem to be getting anything reasonable...
 
blueskadoo42 said:
i have tried setting them equal.
also, tried using the equation X-Xo= Vox*t + .5Ax*t^2
and Vx^2=Vox^2 + 2 Ax (x-xo)

i can't seem to be getting anything reasonable...

How long does it take to reach max height?

Would this be given by Vo/g = t seconds?

Figure that value to decide where the first stone will be when the second is released. (If it has already hit the ground by the time you throw, then it will have already crossed before you threw it. If the first stone is dropping when you throw #2 ... it's good to know what the two are doing so you can keep things straight.
 
blueskadoo42 said:
A stone is thrown vertically upward at a speed of 25.30 m/s at time t=0. A second stone is thrown upward with the same speed 1.860 seconds later. At what time are the two stones at the same height?

what are the equations needed, and any help is greatly appreciated thanks

Write out the time dependence of displacement for both the bodies (keep the 1.860 seconds in mind) and equate them...
 
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