Calculating the Distance of a Ball Kicked Off a Cliff

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A ball weighing 5kg is kicked off a 20-meter cliff, and the goal is to calculate the distance it travels. The gravitational acceleration is 9.80 m/s², and the final velocity (v(f)) is calculated to be 10 m/s, while the initial velocity (v(i)) is 0 m/s. The time of flight is determined to be approximately 2.019 seconds. There is confusion regarding the impulse value of 50 kg m/s, as it was not mentioned in the problem statement. The discussion highlights the need for clarity on the impulse and its relevance to the problem.
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Homework Statement



A ball weighing 5kg is kicked off a cliff with the height of 20 meters.
What is the distance the ball travels?

Known:

g (gravity) =9.80
I (impulse) = 50kg m/s
m (mass) =5kg
h (height) =20m

Unknown:

d=?

Homework Equations



?

The Attempt at a Solution



Found the v(f) which is 10 m/s.
Found the v(i) which is 0 m/s.
 
Last edited:
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t = square root of 2*20/9.81

t= 2.019 seconds
 
RiversAlone said:

Homework Statement



A ball weighing 5kg is kicked off a cliff with the height of 20 meters.
What is the distance the ball travels?

Known:

g (gravity) =9.80
I (impulse) = 50kg m/s
m (mass) =5kg
h (height) =20m
How do you know the impulse is 50 kg m/s? The problem statement did not say anything about impulse.

Unknown:

d=?

Homework Equations



?

The Attempt at a Solution



Found the v(f) which is 10 m/s.
Found the v(i) which is 0 m/s.
How did you get these?
 
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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