Projectiles launched at an angle question

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Homework Help Overview

The problem involves a projectile launched from the ground at a velocity of 20.0 m/s and an angle of 27.0° to the horizontal, with the goal of determining the maximum height reached by the projectile.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve for the maximum height using trigonometric functions and kinematic equations but expresses confusion about the correct method for finding height compared to horizontal range.
  • Some participants question the understanding of vertical velocity at maximum height and suggest considering the components of motion separately.
  • Others raise questions about the kinematic equations applicable to the problem, particularly focusing on the relationship between initial and final velocities, displacement, and acceleration.

Discussion Status

The discussion is ongoing, with participants exploring various aspects of projectile motion. Guidance has been offered regarding the importance of understanding the components of motion and the specific kinematic equations that apply to vertical displacement. There is no explicit consensus yet, as multiple interpretations and methods are being discussed.

Contextual Notes

Participants are navigating the complexities of projectile motion, including the distinction between vertical and horizontal components, and the implications of acceleration due to gravity. There is an emphasis on understanding the reasoning behind each step rather than memorizing procedures.

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Homework Statement

An object is thrown from the ground into the air with a velocity of 20.0m/s at an angle of 27.0° to the horizontal what is the maximum height?

v= 20.0 m/s a=-9.81m/s^2 angle = 27.0

Homework Equations

a=v/t , v= d/t

The Attempt at a Solution

sin(27.0°) = y/20.0m/s (first i solved the vertical direction)

y = sin(27.0°) x 20.0m/s

y =9.07980 m/s

a=v/t (times t then divided by a)

t== vf - vi / a , t=-9.07980 - 9.07980 / -9.81m/s^2

t=1.8511s

cos(27.0) = x/20.0m/s

x = cos(27.0) x 20.0m/s

x= 17.8201m/s

v=d/t , d= VxT

d = 17.8201m/s x 1.8511s

d = 30.0 m ?

I thought you solved it the same way as you solve the object going horizontally. I am really confused how to solve for height.

Thanks
 
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I thought you solved it the same way as you solve the object going horizontally. I am really confused how to solve for height.
The height is not the same as the range - therefore it is not the same method.
You have to think about why you are doing each step and not just memorize the steps.
The height is all about the y component of the motion ... so you have v and angle A.

The y component of the initial velocity is v.sinA ... well done.
The acceleration due to gravity in the +y direction is -g (no need to put numbers in yet)

t== vf - vi / a , t=-9.07980 - 9.07980 / -9.81m/s^2
... what is the y-velocity when the projectile has reached it's maximum height?

Do you know a kinematic equation which will give you the final displacement, given the initial and final velocities, and the acceleration?
 
Simon Bridge said:
... what is the y-velocity when the projectile has reached it's maximum height?

would it be 9.0798m/s?

Do you know a kinematic equation which will give you the final displacement, given the initial and final velocities, and the acceleration?

d=vi.t + 1/2.a.t^2 ?
 
Simon Bridge said:
... what is the y-velocity when the projectile has reached it's maximum height?

SmallPub said:
would it be 9.0798m/s?

If that were true, why isn't the projectile still traveling upward?

If something is not moving in a certain direction, what is its velocity in that direction?
 
would [ y-velocity when the projectile has reached it's maximum height] be 9.0798m/s?
What is the value of y when the ball has that speed??

You need to watch a projectile in motion. Get something you can throw - a ball is good.
Throw it straight up. Catch it. Watch carefully - do it several times.

Observe:
At what y-position is the speed a maximum?
At what y-position is the speed a minimum?
While it is going up, what happens to it's vertical speed?
While it is going down, what happens to it's vertical speed?
When it is at exactly the top of it's trajectory, what is it's speed?

d=vi.t + 1/2.a.t^2 ?
That's a good kinematic equation - but it has time in it.
You don't know the time ... you need one for displacement, given the initial and final velocities and the acceleration.
i.e. you want d, given vi, vf, and a.
 

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