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

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SUMMARY

The speed of a projectile when it leaves a cannon can be calculated using the conservation of energy principle and gravitational potential energy equations. The projectile reaches a height of 35,000 m, and the correct initial speed is determined to be 827 m/s, corresponding to option B. The relevant formula used is v = √(GM/r), where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.

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  • Knowledge of the formula v = √(GM/r)
  • Basic concepts of projectile motion
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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



there's 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}
 

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