Projectile Problem: Stones Meet at Height

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SUMMARY

The problem involves two stones thrown vertically with an initial velocity of 10.5 m/s. The first stone is thrown at time t=0 seconds, while the second stone is thrown 1 second later. To determine the height at which the two stones meet, the kinematic equation s(t) = -4.9t² + v₀t + h is utilized, where v₀ is the initial velocity and h is the initial height. The solution requires calculating the time of flight for both stones and setting their height equations equal to find the meeting point.

PREREQUISITES
  • Kinematic equations for uniformly accelerated motion
  • Understanding of initial velocity and time of flight
  • Basic algebra for solving equations
  • Concept of relative motion in vertical trajectories
NEXT STEPS
  • Study the kinematic equations in detail, especially s(t) = -4.9t² + v₀t + h
  • Learn how to derive the time of flight for projectile motion
  • Explore problems involving multiple objects in motion to enhance understanding
  • Practice solving vertical motion problems with varying initial velocities
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Students studying physics, educators teaching kinematics, and anyone interested in solving projectile motion problems.

Neerolyte
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Suppose you throw a stone straight up with an initial velocity of 10.5 m/s and, 1.0 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?
 
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What have you tried? Where are you stuck?

You need to know the kinematics equations to answer this problem.

HINT: The time for the second stone is (t-1)
 
hm...basically i need help for the whole question
i have no idea how to even start the question
 
s(t)=-4.9t2+v0t+h

Using this and the previous suggestion, you shouldn't have any trouble.
 

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