# Projectile Motion of a Cannonball Problem

• eric_kevin
In summary, the problem involves calculating various parameters of a cannonball launched at a 45 degree angle with a velocity of 2.2 x 10^2 m/s and negligible air resistance. The maximum height can be found using the vertical equations, while the time of flight can be determined by splitting the diagonal velocity into its horizontal and vertical components and using vf = vi + at. The horizontal range can then be calculated using d = 0.5(vi+vf)t, and the final impact velocity can be found using a^2 + b^2 = c^2.
eric_kevin

## Homework Statement

A canon is set at an angle of 45 degrees above the horizontal. A cannonball leaves the muzzle with a speed of 2.2 x 10^2 m/s. Air resistance is negligible. Determine the cannonball's

A)Maximum height
B)Time of Flight *** this is the one I am stuck on
C)Horizontal Range (to the same vertical level)
D)Velocity at Impact

The answers are: A) 1.2 x 10^3 m
B) 32 s
C) 4.9 x 10^3 m
D) 2.2 x 10^2 m

## Homework Equations

vertical equations
v2y^2 = v1y^2 +2ay$$\DeltaD$$
Dy=viy*t+1/2at^2
Dy=vfy*t - 1/2at^2

horizontal equations
vix =dx/t

## The Attempt at a Solution

A) vi y = (sin 45)(220m/s)
= 155.56 m/s

**rearrange to solve for d
vf^2 - vi^2 / 2a= d
0 - (155.56m/s)^2 / -19.6
= 1234.64 m
= 1.2 x 10^3 m

** so finding the maximum height isn't the problem ^
i can't seem to find the time

## Homework Statement

vi x = (cos 45)(220m/s) vi y = (sin45)(220m/s)
= 155.56 m/s = 155.56m/s
t= ? a = -9.8 m/s/s
d=? vf y (at peak) = 0 m/s

You first need to split the diagonal velocity into it's horizontal and vertical components.

Once you know the vertical speed you can use vf = vi + at to calculate the time to the highest point.

Where vf = 0m/s (vertical speed at peak is 0), vi = initial vertical speed, a = -9.8m/s^2

Plug in your values and rearrange to give the time to the peak or maximum height.

Then using d = 0.5(vi+vf)t, you get the maximum height d metres.

For the descent phase, using vf^2 = vi^2 + 2ad, you can insert the initial speed vi (0m/s), the acceleration a (+9.8m/s) and height d (m) you can get the final velocity.

Once you have that, again using vf = vi + at you can get the time of the descent phase.

Add the two times together and you have the full time of the flight.

Because you have split the velocity into it's horizontal and vertical components and now you have the time of the flight, you can use d = 0.5(vi+vf)t to get the horizontal distance. (Ignoring air resistance means there is no deceleration in the horizontal component so vi = vf.)

Again, the velocity in the horizontal direction is constant, so you know that. You also have the final speed from earlier so you just need to combine them using a^2 + b^2 = c^2 to give you the final impact velocity.

See how you go with that, I haven't checked any values so see how you go.

Jared

Last edited:

## 1. How do you calculate the initial velocity of a cannonball?

The initial velocity of a cannonball can be calculated using the formula v = √(g * d), where v is the initial velocity, g is the acceleration due to gravity (usually taken as 9.8 m/s^2), and d is the distance traveled by the cannonball.

## 2. What factors affect the trajectory of a cannonball?

The trajectory of a cannonball is affected by several factors such as the initial velocity, angle of launch, air resistance, and gravity. Other factors that may affect the trajectory include wind speed and direction, temperature, and any external forces acting on the cannonball.

## 3. How does air resistance affect the motion of a cannonball?

Air resistance, also known as drag, slows down the motion of a cannonball. As the cannonball moves through the air, it experiences a force in the opposite direction of its motion. This force increases as the speed of the cannonball increases. Therefore, air resistance can significantly affect the trajectory of a cannonball.

## 4. What is the maximum height reached by a cannonball during its flight?

The maximum height reached by a cannonball can be calculated using the formula h = (v^2 * sin^2θ) / (2g), where h is the maximum height, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity. This formula assumes no air resistance.

## 5. Can a cannonball ever reach its initial height?

In theory, a cannonball can reach its initial height if the angle of launch is set to 90 degrees and there is no air resistance. However, due to the effects of air resistance and other external factors, it is not possible for a cannonball to reach its exact initial height in real-world situations.

Replies
2
Views
1K
Replies
6
Views
534
Replies
11
Views
1K
Replies
16
Views
2K
Replies
4
Views
1K
Replies
18
Views
2K
Replies
11
Views
2K
Replies
11
Views
599
Replies
5
Views
2K
Replies
3
Views
2K