What was the speed of the projectile when it left the cannon?

AI Thread Summary
To determine the speed of a projectile launched from Earth that reaches a height of 35,000 m, the conservation of energy principle is applied. The relevant equation involves gravitational potential energy and kinetic energy, where the initial and final states are considered. The mass of the projectile is not needed since it cancels out in the calculations. The correct speed of the projectile when it left the cannon is identified as 827 m/s. This solution emphasizes the importance of understanding gravitational potential energy in projectile motion.
msattar07
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Homework Statement



A projectile is shot from the surface of te Earth by means of a very powerful cannon. If the projectile reaches a height of 35,000 m above Earth's surface, what was the speed of the projectile when it left the cannon?

a. 355 m/s
b. 827 m/s
c. 710 m/s
d. 906 m/s

Homework Equations



v= square root of GM/r

The Attempt at a Solution



theres no mass, so i wasnt sure how to solve it exactly
 
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What does capital M denote?
 
mass of the Earth
 
Well, that's not an unknown mass.
 
yea so i plugged it in but didnt get the right answer.

the answer is suppose to be B.
 
Well, use conservation of energy:

U_0 + K_0 = U_f + K_f

You know that U = -\frac{GmM}{r}
 
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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