# Calculating the Distance that a Catapult can launch a Projectile

• B
• GopherTv
In summary, the rock is launched from the deck railing and is projected to a certain height and angle.
GopherTv
TL;DR Summary
How do I calculate the distance to which a catapult can launch rocks of different mass?
I know the spring constant of my spring, 90.54 n/m
The spring is stretched 13.5 cm
Im launching of a deck that is 1.75 meters high

How @GopherTv and welcome to PF.

How about measuring the distance? What is calculating going to do for you that measuring won't. If you want to calculate a number using a formula, at the very least you will need to know the speed and the angle at which the rock leaves the catapult. Also, it looks like you are shooting this out a window, so you will need to know the vertical height from the initial position of the rock to the ground.

erobz
GopherTv said:
I know the spring constant of my spring, 90.54 n/m
The spring is stretched 13.5 cm
Calculate the energy stored in the spring that will become kinetic energy.
Weigh the rock, compute the initial velocity for that rock.
Measure the launch angle. Predict the trajectory.

hutchphd
Baluncore said:
Calculate the energy stored in the spring that will become kinetic energy.
Weigh the rock, compute the initial velocity for that rock.
Measure the launch angle. Predict the trajectory.
Should I account for the rotational energy of the lever arm or will that be very little

GopherTv said:
Should I account for the rotational energy of the lever arm or will that be very little
The accuracy of the prediction will be dependent on how much energy accounting you do.
If you avoid computation by experimental calibration, then you could avoid much of the physics. You might then need to use standard rocks.

kuruman said:
Also, it looks like you are shooting this out a window
OP says that it is off a deck. The diagonal members on the bottom which you are taking to be the sill of a window appear to be the railing of a stairway leading down from the deck. He is launching from the deck railing. The corner of the window top that you imagine appears to be the curve of the downspout for the roof gutter.

Summing the torques about the point I forgot to label ( also I didn't include the force of the weight of the rod in the diagram ):

$$b F_s \sin \beta - r m g \cos \theta - \frac{r}{2} m_r g \cos \theta = \left( I_{r_G} + m_r \left( \frac{r}{2} \right)^2 + m r^2 \right) \frac{ d \omega }{dt}$$

Where;

## I_{r_G} ## is the mass moment of inertia (M.o.I) of the rod about its CM
## m_r ## is the mass of the rod
$$F_s = k ( x - x_o )$$

$$\sin \beta = \frac{ \sqrt{d^2 + c^2} \sin \varphi }{x}$$

$$\varphi = \pi - \left( \theta + \arctan \left( \frac{d}{c} \right) \right)$$

and

$$x = \sqrt{ d^2 + c^2 + b^2 - 2 b \sqrt{d^2 + c^2} \cos \varphi }$$

You'll want to integrate this numerically to determine the angular velocity until the angel of launch (it should be close to ## \pi/2 ## ), and then extract the tangential velocity from:

$$v = r \omega$$

From there its kinematics (ignoring air resistance )
Like I said, not an analytical result, but you can program it without much difficulty. If you have questions please feel free to comment.

Last edited:

## 1. How is the distance that a catapult can launch a projectile calculated?

The distance that a catapult can launch a projectile is calculated using the formula: d = v2sin(2θ)/g, where d is the distance, v is the initial velocity of the projectile, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s2).

## 2. What factors affect the distance that a catapult can launch a projectile?

The distance that a catapult can launch a projectile is affected by several factors such as the initial velocity of the projectile, the launch angle, the weight and size of the projectile, and the strength and design of the catapult.

## 3. How does the launch angle affect the distance that a catapult can launch a projectile?

The launch angle plays a significant role in determining the distance that a catapult can launch a projectile. A higher launch angle will result in a longer distance, as the projectile will have a higher vertical component of velocity, allowing it to stay in the air for a longer time.

## 4. Can the distance that a catapult can launch a projectile be increased?

Yes, the distance that a catapult can launch a projectile can be increased by adjusting the launch angle, using a more powerful catapult, or using a lighter and more aerodynamic projectile. However, there is a limit to how far a catapult can launch a projectile, as it is limited by factors such as air resistance and the strength of the catapult itself.

## 5. How accurate is the calculation of the distance that a catapult can launch a projectile?

The calculation of the distance that a catapult can launch a projectile is an idealized calculation and may not always accurately represent the actual distance achieved. Factors such as air resistance, wind, and imperfections in the catapult can affect the actual distance. Therefore, the calculated distance should be considered as an estimate rather than an exact value.

### Similar threads

• Classical Physics
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
137
• Classical Physics
Replies
4
Views
115
• Engineering and Comp Sci Homework Help
Replies
1
Views
375
• Introductory Physics Homework Help
Replies
38
Views
2K
• Classical Physics
Replies
14
Views
913
• Classical Physics
Replies
1
Views
814
• Classical Physics
Replies
2
Views
970
• Classical Physics
Replies
27
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
663