How to find the Eg of an object thrown at an angle?

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In summary, the balls have the same energy at the start, but have different speeds when they hit the ground.
  • #1
nilfound
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Homework Statement


Four identical balls are thrown from the top of a cliff, each with the same speed. The first is thrown straight up, the second is thrown at 30° above the horizontal, the third at 30° below he horizontal, and the fourth straight down. How do the speeds and kinetic energies of the balls compare as they strike the ground…
  1. when air resistance is negligible?
  2. when air resistance is significant?

Homework Equations


Eg = mgh, Ek = 1/2mv^2
Et = Ek where height <= 0
Et = Eg where height > 0

The Attempt at a Solution


Eg = mghcos()?
 
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  • #2
All the balls start with the same speed from the same height, so have the same energy at that point.
 
  • #3
But when they hit the ground, the angle they were thrown at will have a significant difference in their speed as gravity grabs hold right? So won't the first ball, in the extreme case of gained height, have a greater final velocity and kinetic energy? Or am I missing a fundamental part of energy conservation :p
 
  • #4
Because there are two heights: the starting height at the top of the cliff, and the ending height at the bottom. I am just confused on how to calculate how they will be different or the same!
 
  • #5
For energy it will be the same of course! But the velocities will vary will they not? But then if the velocity varies, then Ek changes, because Ek = 1/2mv^2!
 
  • #6
Velocity can have a vertical and a horizontal component.
 
  • #7
nilfound said:
How do the speeds and kinetic energies of the balls compare as they strike the ground…
Is that the exact wording of the question? Or is it maybe asking about the vertical component of the velocity when the balls hit the ground?
 
  • #8
Also, can you list the relevant kinematic equations of motion for a constant acceleration (gravitational acceleration downward)? Those will be helpful in solving this problem. :smile:
 

FAQ: How to find the Eg of an object thrown at an angle?

How do you calculate the initial velocity of an object thrown at an angle?

To calculate the initial velocity (v0) of an object thrown at an angle, you can use the formula: v0 = v * cos(θ), where v is the velocity of the object and θ is the angle at which it was thrown.

How do you find the horizontal and vertical components of the initial velocity?

The horizontal component (v0x) of the initial velocity can be calculated using the formula: v0x = v * cos(θ), where v is the velocity of the object and θ is the angle at which it was thrown. The vertical component (v0y) can be calculated using the formula: v0y = v * sin(θ).

What is the formula for finding the range of an object thrown at an angle?

The range (R) of an object thrown at an angle can be calculated using the formula: R = (v02 * sin(2θ)) / g, where v0 is the initial velocity of the object, θ is the angle at which it was thrown, and g is the acceleration due to gravity.

How does air resistance affect the motion of an object thrown at an angle?

Air resistance can cause a decrease in the horizontal distance and an increase in the time of flight of an object thrown at an angle. This is because air resistance acts against the motion of the object, slowing it down and reducing its range. However, the effect of air resistance is more significant at higher velocities and can be negligible for objects thrown at lower angles and speeds.

Can the launch angle affect the maximum height of an object thrown at an angle?

Yes, the launch angle can affect the maximum height of an object thrown at an angle. The maximum height (H) can be calculated using the formula: H = (v02 * sin2(θ)) / (2g), where v0 is the initial velocity of the object, θ is the launch angle, and g is the acceleration due to gravity. As the launch angle increases, the maximum height also increases, reaching its maximum at a 90-degree angle (vertical launch).

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