What are some tips for understanding projectiles in physics?

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In summary, the conversation was about a student struggling with projectiles in their physics course and seeking advice on how to better understand them. They received tips on reading the text carefully and asking specific questions, as well as explanations on vectors, components of vectors, and equations for finding position in horizontal and vertical parts. The student later thanked for the help and mentioned having a better understanding.
  • #1
hasnain721
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Hi there,
I am doing physics as level and i am finding it very difficult especially the projectiles. I am unable to say when u=0 or v=0 and i am not sure how to analyse the situation vertically and horizontally!

Some one please suggest me a good revision guide. I am following the edexcel course (not the salters one).


Plz reply
Thanks.
Hasnain Mir Mohammed
 
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  • #2
Hi, hasnain!

First tip is to read the text CAREFULLY. That is where the clues are given to you.
You can also "ask" questions to the text, like re-reading it with specific questions in your mind like "What does this text say about the horizontal velocity component?".

Having such specific questions in your head while reading the text will alert your consciousness to those passages containing the answer to your question.

As for decomposed analysis, always ask yourself:
"What are the forces I'm asked to take into account?"
Then, having identified those, ask yourself "in which directions do these forces act?".
 
  • #3
Hi dear,

May be i can help u with this...with reasoning...

First of ll u should know about Vectors.

Vectors are Those quantities which show their effect in a direction...ok

now our purpose is to find position of particlee at various intervals

now,we should know two equations for purpose of calculation...

These are-
1.[in case velocity (value as well as direction) is not changing]
X= x(at starting) + v.t

remember,X =final position, x= initial position and v=velocity
this equation cannot be applied until we get a velocity having fixed direction and fixed value

2.[for constant acceleration (in direction as well as value)]

X= x + ut + 1/2 a t²

this equation also can't be applied in circular motion


Now,for calculation of position,we use above two equations or the equation for position as given in question (or as studied in our textbooks...for example - for circular motion,we know the equation for position already)

NOW WE NEED TO KNOW JUST ONE MORE THING- COMPONENTS OF VECTOR

-As we know vector have their effect in one direction,this effect can be observed in parts also-for example,an object moving in NORTH-EAST direction goes in NORTH as well as EAST

The components of a vector are equal to

v.cosӨ and v.sinӨ in a plane

so we can find position in PARTS and ADD them together to find total value..

now in projectile,

Gravity acts downwards (vertically),so there is NO GRAVITY IN HORIZONTAL DIRECTION.

now,to find positions in projectile,we divide velocity and gravity(gravitational acceleration) in HORIZONTAL and VERTICAL PARTS(or components)

we get,

IN HORIZONTAL...

VELOCITY=V.COSӨ
GRAVITATIONAL ACCELERATION(GRAVITY) = 0 (as told u above)

IN VERTICAL

VELOCITY = V.SINӨ
GRAVITY = g (=9.8)

now we use FOLLOWING EQUATIONS(as told u above) for finding position in HORIZONTAL and VERTICAL parts

1.FOR HORIZONTAL
X= x(at starting) + v.t (Because here velocity is fixed as no acceleration is there)

and we know v=horizontal velocity=V.COSӨ

2.FOR VERTICAL
X= x + ut + 1/2 a t² (Because here acceleration is fixed)

and we know v=vertical velocity=V.SINӨ

NOTE-
(1) u=o if object is not moving in starting (like when object is dropped v.cosӨ and v.sinӨ both are =0 but when object is thrown parallel to ground,only v.cosӨ is =0,v.sinӨ is not equal to 0)

(2) v= 0 if object stops at end (like when object reaches to its highest position,v.sinӨ is =0 but v.cosӨ is not equal to 0)

i hope u will get all these.reply me..bye
 
  • #4
mr.survive said:
Hi dear,

May be i can help u with this...with reasoning...

First of ll u should know about Vectors.

Vectors are Those quantities which show their effect in a direction...ok

now our purpose is to find position of particlee at various intervals

now,we should know two equations for purpose of calculation...

These are-
1.[in case velocity (value as well as direction) is not changing]
X= x(at starting) + v.t

remember,X =final position, x= initial position and v=velocity
this equation cannot be applied until we get a velocity having fixed direction and fixed value

2.[for constant acceleration (in direction as well as value)]

X= x + ut + 1/2 a t²

this equation also can't be applied in circular motion


Now,for calculation of position,we use above two equations or the equation for position as given in question (or as studied in our textbooks...for example - for circular motion,we know the equation for position already)

NOW WE NEED TO KNOW JUST ONE MORE THING- COMPONENTS OF VECTOR

-As we know vector have their effect in one direction,this effect can be observed in parts also-for example,an object moving in NORTH-EAST direction goes in NORTH as well as EAST

The components of a vector are equal to

v.cosӨ and v.sinӨ in a plane

so we can find position in PARTS and ADD them together to find total value..

now in projectile,

Gravity acts downwards (vertically),so there is NO GRAVITY IN HORIZONTAL DIRECTION.

now,to find positions in projectile,we divide velocity and gravity(gravitational acceleration) in HORIZONTAL and VERTICAL PARTS(or components)

we get,

IN HORIZONTAL...

VELOCITY=V.COSӨ
GRAVITATIONAL ACCELERATION(GRAVITY) = 0 (as told u above)

IN VERTICAL

VELOCITY = V.SINӨ
GRAVITY = g (=9.8)

now we use FOLLOWING EQUATIONS(as told u above) for finding position in HORIZONTAL and VERTICAL parts

1.FOR HORIZONTAL
X= x(at starting) + v.t (Because here velocity is fixed as no acceleration is there)

and we know v=horizontal velocity=V.COSӨ

2.FOR VERTICAL
X= x + ut + 1/2 a t² (Because here acceleration is fixed)

and we know v=vertical velocity=V.SINӨ

NOTE-
(1) u=o if object is not moving in starting (like when object is dropped v.cosӨ and v.sinӨ both are =0 but when object is thrown parallel to ground,only v.cosӨ is =0,v.sinӨ is not equal to 0)

(2) v= 0 if object stops at end (like when object reaches to its highest position,v.sinӨ is =0 but v.cosӨ is not equal to 0)

i hope u will get all these.reply me..bye

Hey!
Sorry for the late reply as i didnt know that u had posted!

I have now developed a better understand of projectiles. I was just panicking.

Thanks for the detailed post.
 

1. What are projectiles?

Projectiles are objects that are launched or thrown and move through the air under the force of gravity. Examples include bullets, baseballs, and arrows.

2. How do projectiles move?

Projectiles move in a curved path, known as a parabola, due to the downward acceleration of gravity. The initial force of the launch and the angle at which the projectile is launched also affect its motion.

3. How is the trajectory of a projectile calculated?

The trajectory of a projectile can be calculated using mathematical equations and principles, including Newton's laws of motion and the equations of motion for projectile motion. It also takes into account variables such as initial velocity, launch angle, and the force of gravity.

4. What factors affect the range of a projectile?

The range of a projectile is affected by the initial velocity, launch angle, and the force of gravity. Other factors such as air resistance, wind, and the shape and weight of the projectile can also impact its range.

5. How can projectile motion be applied in real life?

Projectile motion is a fundamental concept in physics and has many practical applications, such as in sports, military operations, and space exploration. Understanding projectile motion can also help engineers design and improve technologies such as rockets and cannons.

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