How High Does a Coin Tossed Straight Up Reach?

AI Thread Summary
A coin tossed straight up takes 2.75 seconds to return to its starting point, which is 1.30 meters above the ground. To find the maximum height, the initial velocity (V0) can be calculated using the formula V = V0 + gt, where V is zero at the peak height. The maximum height can then be determined using the equation V^2 = V0^2 + 2g(ay), where ay is the additional height gained. The solution involves combining these equations to find the total height reached by the coin. Understanding these physics principles is essential for solving the problem effectively.
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Homework Statement


A coin tossed straight up into the air takes
2.75 s to go up and down from its initial
release point 1.30 m above the ground.

What is its maximum height?


Homework Equations




The Attempt at a Solution





I seriously have no clue how to do this problem. I really need a detailed explanation with steps. I'm sorry if this doesn't abide by your question standards, but I really mean I have 0 idea how to solve this problem, and I really need some help.


Thanks.
 
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can anyone help?
 
its 1.30 + V0^2/2g

since V^2=V0^2 + 2g(ay) where the max height is when V=0
 
where V0=g(2.75/2) since V= V0 + gt where V=0m/s since you are accounting for the time it takes and the velocity at the point it goes its max height.
 
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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