Simple projectile motion, yet stuck.

AI Thread Summary
A student threw a ball horizontally from a height of 8.0m and it was caught 10.0m away, prompting a calculation for the initial velocity. The relevant equations for projectile motion were discussed, including the time of flight derived from the height. The calculated time for the ball to fall was found to be approximately 1.63 seconds. The initial velocity was determined to be 7.16 m/s based on the horizontal distance covered. Clarification was sought regarding the height at which the ball was caught and the position of the other student.
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Homework Statement


A student threw a ball horizontally out a window 8.0m above the ground. It was caught by another student 10.0m away. What was the initial velocity of the ball?
ANSWER PROVIDED: 7.16m/s

Homework Equations


dv=1/2at^2
t=vv/a
dh=vht

The Attempt at a Solution


dv=1/2at^2
8=1/2(9.8)t^2
√ans
t=1.63sec

The fact that I'm trying to find it's initial velocity causes my brain to become a blank slate when deciding on what my next step should be (if the first is even right).
 
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is the other student supposed to be at ground level or something? What's the height which the ball is caught?
 
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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