How High Was the Basketball Dropped to Achieve a Momentum of 2.5 kgm/s?

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To determine the height from which a basketball was dropped to achieve a momentum of 2.5 kgm/s, the mass of the basketball (0.57 kg) is used to calculate its velocity. The momentum formula (mass * velocity) confirms that the velocity can be found by dividing momentum by mass. The kinematic equation (vf^2 = vi^2 + 2ah) is then applied to relate velocity, acceleration, and height. By solving for velocity first and then using the kinematic equation, the height can be calculated. This approach effectively guides the problem-solving process for the physics question.
TastyTyr
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Ok, I've done work on this problem, but I think that everything I try is wrong. I can't get any help from my book or my teacher because he made it up and wants up to try it. But I have no clue how to start. Please help me :cry:

A basketball (m = 0.57 kg) is dropped from rest. Just before striking the floor, the magnitude of the basketball's momentum is 2.5 kgm/s. At what height was the basketball dropped?
 
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mass * velocity = momentum right?
you know the mass so you can find velocity.
do you know any equations that include velocity, acceleration and height?
 
Oh yeah...and so I would solve for velocity first then I would use the kinematic equation,for velocity acceleration and height? (vf^2)=(vi^2)+2ah ??
 
you got it
 
Thanks a bunch!:smile:
 
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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