When faced with a projectile motion problem, there are things you must do and remember.
Do
1. determine the horizontal and vertical components of the initial velocity.
For an object projected horizontally, the vertical component of initial velocity is zero.
For an object thrown at 20 m/s 53º above the horizontal, the vertical component of the initial velocity is 20sin53º = 16.0 m/s [up], and the horizontal component is 20cos53º = 12.0 m/s.
2. break down the problem into two problems: horizontal and vertical.
3. assign - and + signs appropriately
Remember:
1. acceleration is zero in the horizontal component and g in the vertical.
2. at the top of its rise, the projectile has a vertical velocity = zero.
(For the two types of projectile motion below, I had to defy Mr. Chung's use of certain letters for certain values because the diagrams that I found used different letters.)
TYPE 1
x-components
1. Component of initial velocity along x-axis.
ux=u
ux=u
2. Acceleration along x-axis
ax=0
(Because no force is acting along the horizontal direction)
ax=0
(Because no force is acting along the horizontal direction)
3. Component of velocity along the x-axis at any instant t.
vx=ux + axt
=u + 0
» vx=uvx=ux + axt
=u + 0
This means that the horizontal component of velocity does not change throughout the projectile motion.
4. The displacement along x-axis at any instant t
x=uxt + (1/2) axt2
» x=uxt + 0
» x=u t
y-components
1. Component of initial velocity along y-axis.
uy=0
2.Acceleration along y-axis
g=9.8m/s2
3. Component of velocity along the y-axis at any instant t.
vy=uy + ayt
=0 + gt
» vy=gt
4. The displacement along y-axis at any instant t
y= uyt + (1/2) ayt2
» y= 0 + (1/2) ayt2
» y=1/2gt2
TYPE 2
x-components
1. Component of initial velocity along x-axis.
ux=u cosØ [Ø = theta]
2. Acceleration along x-axis
ax=0
(Because no force is acting along the horizontal direction)
ax=0
(Because no force is acting along the horizontal direction)
3. Component of velocity along the x-axis at any instant t.
vx=ux + axt
=ucosØ + 0
vx=ux + axt
=ucosØ + 0
= ucosØ
» vx=ucosØThis means that the horizontal component of velocity does not change throughout the projectile motion.
4. The displacement along x-axis at any instant t
x=uxt + (1/2) axt2
» x=(ucosØ)t
y-components
1. Component of initial velocity along y-axis.
uy=u sinØ
2.Acceleration along y-axis
g= -9.8m/s2
3. Component of velocity along the y-axis at any instant t.
vy=uy + gt
vy=usinØ + gt
4. The displacement along y-axis at any instant t
y= uyt + (1/2) gt2
» y= (usinØ)t + (1/2)gt2
This is a very useful post especially for the students who are studying this topic. It explains everything so easily and clearly. Thanks for sharing the information with us!
ReplyDeleteThank you so much for making this list.This was extremely helpful!
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