Projectile Problem: Stones Meet at Height

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To solve the projectile problem, the kinematics equations are essential, particularly the equation s(t) = -4.9t² + v₀t + h. The first stone is thrown with an initial velocity of 10.5 m/s, and after 1 second, the second stone is thrown with the same velocity. The time for the second stone's motion is t - 1 seconds, where t is the total time since the first stone was thrown. By setting the heights of both stones equal, one can determine the height at which they meet above the release point. Understanding these concepts will facilitate finding the solution to the problem.
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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