What Is the Speed of a Cannonball at Its Maximum Height?

AI Thread Summary
At maximum height, the vertical velocity of a cannonball is zero, meaning it only retains horizontal velocity. The problem cannot be solved with the provided information, as the initial speed or angle of launch is necessary to determine the horizontal component of velocity. The calculations attempted by the user were based on gravitational force and height but did not account for the initial conditions required. Participants confirmed the need for additional data to accurately solve the problem. Clarification from the teacher is suggested to ensure all necessary information is available.
Chaalie
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Hey guys!
This is my first post on this forum, though I've been looking through it for a few months now. I am in a basic Grade 11 physics course, and I'm having a little trouble with this one. So, here goes...

Homework Statement


A cannonball (mass=15.5kg) is shot out of a cannon on Earth (assume it is shot at a height of 0m). The maximum height that it reaches is 20.1m. What is the speed of the cannonball at the maximum height?

Homework Equations


The different equations I know that might be useful are...
Eg=m*g*h
V=d/t
A=v/t
d=V(initial)*t + (A*(t)^2)/2
d=[(V(initial)+V(final))*t]/2

The Attempt at a Solution


My attempts have been a little rough, but here is my best one...
F=mg
=15.5*9.8
=151.9N

A=F/m
=151.9N/15.5
=9.8m/s^2

Vf^2=Vi^2 + 2A*d
= 0 + 2(9.8)(20.1)
Vf^2= 393.96
Therefore, Vf = 19.84842563m/sI have a feeling I've done something wrong...any input?
I'd still like to figure it out myself, so if somebody could just point me in the right direction, that'd be great!Charlie
 
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The the maximum height, its vertical velocity is zero. It has only horizontal velocity, and horizontal velocity depends on the velocity at which it was fired. It isn't possible to solve this problem with the information you are given.
 
cavalier said:
It isn't possible to solve this problem with the information you are given.

Correct. We need to know the initial speed or angle, or the initial horizontal component of velocity, to solve the problem.
 
Thanks guys!
I'll check with my teacher to see if he missed anything.
 
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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