Solve 2D Kinematics Problem: Max Height & Range

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SUMMARY

The discussion focuses on solving a 2D kinematics problem involving a projectile launched from a cannon at an angle of 30 degrees with an initial speed of 112.7 m/s. The maximum height achieved by the projectile is calculated using the formula h = u²sin²θ/2g, resulting in a height of 120.3 meters. The range of the projectile is determined using the formula R = u²sin2θ/g, yielding a distance of 641.5 meters before it hits the ground. These calculations illustrate the application of kinematic equations in projectile motion.

PREREQUISITES
  • Understanding of basic trigonometry (sine and cosine functions)
  • Familiarity with kinematic equations in physics
  • Knowledge of projectile motion concepts
  • Basic understanding of gravitational acceleration (9.8 m/s²)
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  • Study the derivation and application of kinematic equations for projectile motion
  • Learn how to decompose vectors into horizontal and vertical components
  • Explore the effects of different launch angles on projectile range and height
  • Investigate real-world applications of projectile motion in sports and engineering
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Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion and its calculations.

Lyzriel
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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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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.
 


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!
 

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