# A cannon fires a cannonball 500m downrange when set at a 45 degree angle. At what velocity does the cannonball leave the cannon?

We can write that there will be zero net vertical displacement during this time if T is the flight time.

During this period, the horizontal displacement, or range, will be

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To find the velocity of the cannonball leaving the cannon, we can use the kinematic equation for projectile motion:

[ v = \sqrt{\frac{2gh}{\sin\theta}} ]

Where:

- ( v ) is the initial velocity of the cannonball,
- ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )),
- ( h ) is the maximum height of the projectile (which occurs at half the total flight time),
- ( \theta ) is the launch angle (45 degrees in this case).

Given that the range (( R )) of the projectile is 500 meters and the launch angle is 45 degrees, the maximum height (( h )) can be calculated using the formula:

[ R = \frac{v^2 \sin 2\theta}{g} ]

Solving for ( v ):

[ v = \sqrt{\frac{Rg}{\sin 2\theta}} ]

Substituting the given values:

[ v = \sqrt{\frac{(500 , \text{m})(9.81 , \text{m/s}^2)}{\sin(2 \times 45^\circ)}} ]

[ v = \sqrt{\frac{4905 , \text{m}^2/\text{s}^2}{\sin(90^\circ)}} ]

[ v = \sqrt{\frac{4905 , \text{m}^2/\text{s}^2}{1}} ]

[ v = \sqrt{4905 , \text{m}^2/\text{s}^2} ]

[ v \approx 70.05 , \text{m/s} ]

Therefore, the velocity of the cannonball leaving the cannon is approximately ( 70.05 , \text{m/s} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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