Analyzing Canon Ball's Height & Acceleration Over Time

AI Thread Summary
The discussion focuses on the projectile motion of a cannonball fired at different angles (30°, 45°, and 60°). Participants emphasize that the only force acting on the projectile after launch is gravity, resulting in a constant downward acceleration of -9.8 m/s². They clarify that the vertical velocity reaches zero at the midpoint of the flight time, which is determined by the initial velocity components. It is noted that a greater launch angle increases the time the projectile remains in the air and its maximum height. Understanding these principles allows for accurate graphing of height and acceleration over time.
jadiepants
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Homework Statement



A canon ball is fired from a canon upwards at an angle above the horizontal.

a. Sketch a graph showing the height of this canon ball over time if it was fired at 30°, 45° and 60° above the horizontal.
b. Sketch showing the acceleration of the canon ball over time if it was fired at 30°, 45° and 60° above the horizontal.
c. At what point in the flight of the canon ball is the velocity at a minimum?

Homework Equations


h*sin(angle) = hy
h*cos(angle) = hx

The Attempt at a Solution



I have sketched 3 triangles but how do I work out the components with just an angle? Or is there another way of doing this?
 
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- projectile motion isn't triangles
- once launched, what's the only force on the projectile? therefore the acceleration is?
- at what point in time will the vertical velocity be zero?
 
Ush said:
- projectile motion isn't triangles
- once launched, what's the only force on the projectile? therefore the acceleration is?
- at what point in time will the vertical velocity be zero?

- I'm under the impression that a canon ball being fired is projectile

- the acceleration in the y direction is -9.8, but what use is that when I don't know the initial velocity or how high it goes?

- I don't know when it will be zero because I don't have any other information.

If I'm doing it wrong please just say so.
 
jadiepants said:
- I'm under the impression that a canon ball being fired is projectile

- the acceleration in the y direction is -9.8, but what use is that when I don't know the initial velocity or how high it goes?

- I don't know when it will be zero because I don't have any other information.

If I'm doing it wrong please just say so.

-Therefore your acc'n -time graph will be a straight line.
-A greater angle (from 0 - 90) represents a greater time in the air. (as well as greater height achieved) (why?)
-therefore, you can use relative times.

-you can assume each projectile has the same initial velocity, Vo because it was fired from the same cannon
-vertical velocity is zero at 1/2t
 
thank you so much :)
 
another hint:

If you fire a cannon ball with the following velocities for ex:
Vy = 10, Vx = 10
Vy = 10, Vx = 100
Vy = 10, Vx = 10000
ALL will fall to the ground at the SAME time.

You're Vy, however, is changing because the cannon fires at the same initial velocity. You have to break that initial velocity into its components. You can safely ignore the x-component. A greater angle = A greater Vy component = A greater time in the air.

hope that helps
~Ush
 
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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