How can I solve this basic velocity problem involving a thrown ball?

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To solve the velocity problem involving a thrown ball, the starting velocity can be calculated using kinematic equations based on the height of 12.0 m and the time of 1.25 s to reach the ground. The horizontal velocity can be determined by dividing the horizontal distance of 40.0 m by the time of flight. Additionally, the launch velocity's speed and direction can be found by combining the vertical and horizontal components of the motion. Participants are encouraged to show their calculations and identify where they encounter difficulties for better assistance. Engaging with the problem-solving process is essential for understanding the concepts involved.
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Homework Statement



A ball is thrown from a height of 12.0 m above the ground level. If it takes 1.25 s to reach the ground:

a) Calculate its starting velocity
b) If the ball lands 40.0 m horizontally out from the throwing point, what horizontal velocity did it have?
c) Determine the speed and direction of the launch velocity
 
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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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