# Projectile motion cannonball problem

In summary, the conversation discusses the calculation of various parameters for a fired cannonball, including the time in air, initial velocity, range and height after 18 seconds, and the range at which the height is the same as calculated previously. The equations used involve the initial velocity, angle of elevation, and acceleration due to gravity. The conversation also mentions an error in the analysis of the y component and suggests using the formula s = ut + \frac{1}{2}at^2 or the formula R = \frac{u^2 \sin 2\theta}{g}.

## Homework Statement

A cannonball is fired with an angle of elevation of 70o. Its range is 1km. Neglect the air resistance and calculate:
1. the time the cannonball is in the air.
2. its initial velocity.
3. its range and height at 18s after firing
4. the other range at which the height was the same as calculated in 3 above.

## Homework Equations

u = the initial velocity
ux = ucos70o
uy = u sin70o
a = -9.81m/s2

## The Attempt at a Solution

My attempt.

Range = ucos70o x t
therefore
t = Range/ucos70o

At full range for the Y component.
t = (Vy - u sin70o)/-9.81
t = u sin70o)/9.81

Since the time are same.

Range/ucos70o = u sin70o/-9.81 ( Is this right?)

I am confused. Can someone give me hints...thanks in advance..

You have an error for your analysis of the y component. You haven't justified by V_y is zero at full range, and in fact it isn't! Think about it.

To continue, you should use $$s = ut + \frac{1}{2}at^2$$.

Or if you remember the formula $$R = \frac{u^2 \sin 2\theta}{g}$$ that could be very useful.

I would first clarify any uncertainties or confusion in the problem statement. It is important to have a clear understanding of the given information and what is being asked in order to solve the problem accurately.

In this case, the problem states that the range is 1km, but it is not specified if this is the maximum range or the total distance traveled by the cannonball. This information could affect the calculation of the initial velocity and the other range at which the height is the same.

Assuming that the range given is the maximum range, here is my response to the problem:

1. To calculate the time the cannonball is in the air, we can use the formula for the horizontal distance traveled: R = utcosθ, where R is the range, u is the initial velocity, t is the time, and θ is the angle of elevation. Rearranging the formula, we get t = R/(u cosθ). Substituting the given values, we get t = 1km/(u cos70o).

2. To calculate the initial velocity, we can use the formula for the vertical distance traveled: h = utsinθ - 1/2gt^2, where h is the height, u is the initial velocity, t is the time, and g is the acceleration due to gravity. At the maximum range, the height is 0, so we can solve for u. Rearranging the formula, we get u = √(2gh)/(tsinθ). Substituting the given values, we get u = √(2(9.81m/s^2)(0.001km))/(t sin70o).

3. To calculate the range and height at 18 seconds after firing, we can use the equations for the horizontal and vertical distances traveled, with t = 18s. For the horizontal distance, we get R = utcosθ = (u cos70o)(18s). For the vertical distance, we get h = utsinθ - 1/2gt^2 = (u sin70o)(18s) - 1/2(9.81m/s^2)(18s)^2.

4. To find the other range at which the height is the same as calculated in part 3, we can set the two height equations equal to each other and solve for R. We get R = (u cos70o

## 1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched or thrown into the air and moves along a curved path due to the influence of gravity.

## 2. How is projectile motion affected by air resistance?

Air resistance, also known as drag, can affect the trajectory of a projectile by slowing it down and changing its direction. This is why objects with different shapes and surface areas experience different levels of air resistance.

## 3. What is the equation for calculating the horizontal distance of a projectile?

The equation for calculating the horizontal distance of a projectile is d = v0 * t * cosθ, where v0 is the initial velocity, t is the time of flight, and θ is the launch angle.

## 4. What is the maximum height reached by a projectile?

The maximum height reached by a projectile can be calculated using the equation h = (v02 * sin2θ) / (2 * g), where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 5. How does changing the initial velocity and launch angle affect the trajectory of a projectile?

Changing the initial velocity and launch angle can significantly impact the trajectory of a projectile. A higher initial velocity will result in a longer horizontal distance and a greater launch angle will result in a higher maximum height. Both changes can also affect the time of flight and impact location of the projectile.

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