Solve 2D Kinematics Problem: Max Height & Range

In summary, the conversation discusses the calculation of the maximum height and range of a projectile launched from a cannon at a 30-degree angle with an initial speed of 112.7m/s. The process involves sketching a picture of the situation, decomposing the initial velocity into its horizontal and vertical components, and setting up the value of x and y as a function of time.
  • #1
Lyzriel
1
0
HELP! i don't undersatnd any of it..

like say a canon fires a ball a groiund level at 30deg above the horiz. and the initial speed is 112.7m/s. Whas the max height and the range of the proectile??

None of this is making sense to me...
 
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  • #2
Sketch a picture of the situation.
Then decompose the initial velocity into its horizontal and vertical components.
Since only gravity is accelerating the particle downwards, the horizontal component has uniform velocity. The vertical component is in freefall.
So you can setup up the value of x and y as a function of t.
 
  • #3


Hi there,

I understand that kinematics can be a challenging topic to grasp at first. Let's break down the problem step by step to help you understand it better.

Firstly, we are given the initial angle of the canon, which is 30 degrees above the horizontal. This means that the ball is being launched at an upward angle, rather than straight ahead. The initial speed of the ball is also given, which is 112.7m/s.

Now, to solve for the maximum height and range, we need to use the equations of kinematics. These equations relate the position, velocity, acceleration, and time of an object in motion.

For the maximum height, we need to use the equation h = u^2sin^2θ/2g, where h is the maximum height, u is the initial speed, θ is the angle of launch, and g is the acceleration due to gravity (9.8m/s^2). Plugging in the values given in the problem, we get:

h = (112.7m/s)^2sin^2(30deg)/2(9.8m/s^2) = 120.3m

Therefore, the maximum height reached by the ball is 120.3m.

To solve for the range, we can use the equation R = u^2sin2θ/g, where R is the range. Plugging in the values, we get:

R = (112.7m/s)^2sin(2(30deg))/9.8m/s^2 = 641.5m

This means that the ball will travel a horizontal distance of 641.5m before hitting the ground.

I hope this explanation helps you understand the problem better. Remember to always write down the given values and use the appropriate equations to solve for the unknowns. If you have any further questions, please don't hesitate to ask. Good luck!
 

Related to Solve 2D Kinematics Problem: Max Height & Range

What is 2D kinematics and how is it different from 1D kinematics?

2D kinematics is the study of motion in two dimensions, typically involving the x and y axes. It differs from 1D kinematics, which only considers motion along a single axis, typically the x axis.

What is the equation for calculating the maximum height of an object in 2D kinematics?

The equation for calculating the maximum height of an object in 2D kinematics is h = (v02sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

How do you find the range of an object in 2D kinematics?

The range of an object in 2D kinematics can be found using the equation R = (v02sin2θ)/g, where R is the range, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

Can you use the same equations for 2D kinematics on Earth and other planets?

The equations for 2D kinematics can be used on any planet as long as the acceleration due to gravity (g) is known for that planet. However, other factors such as air resistance and different gravitational forces may affect the motion of an object on different planets.

How can 2D kinematics be applied in real-life situations?

2D kinematics can be applied in real-life situations such as calculating the trajectory of a projectile, predicting the motion of objects in sports like football or basketball, and designing roller coasters and other amusement park rides.

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