moving in the x and y direction parabola curve.parabola (caused by both horizontal motion and...
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![Page 1: Moving in the x and y direction parabola curve.parabola (caused by both horizontal motion and vertical motion. It must accelerate only in the vertical direction) * * The projectile](https://reader033.vdocument.in/reader033/viewer/2022060816/6095445398f08177812cc7ce/html5/thumbnails/1.jpg)
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*Moving in the x and y direction
*A projectile is an object shot
through the air. This occurs in a
parabola curve.
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Object
droppedObject
thrown up
Object thrown at an
angle
projectile- any object that moves through the air
or through space, acted on only by gravity
(and air resistance, if any)
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The vertical acceleration of a
projectile is caused by gravity, so
ay = -9.8 m/s2
Parabolic
Trajectory
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*g remains constant (g= -9.8m/s2)
*a in the x direction is 0 because gravity is not
acting on it.
*Neglect air resistance
*Neglect the effects of the earths rotation
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Projectiles launched horizontally
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To find how far the ball falls, you
use the formula. y =viyt + 1/2gt2
1st second- 5m
After 2 seconds- 20m
After 3 seconds- 45m
The curved path of a
projectile produced is a
parabola (caused by both
horizontal motion and vertical
motion. It must accelerate
only in the vertical direction)
![Page 8: Moving in the x and y direction parabola curve.parabola (caused by both horizontal motion and vertical motion. It must accelerate only in the vertical direction) * * The projectile](https://reader033.vdocument.in/reader033/viewer/2022060816/6095445398f08177812cc7ce/html5/thumbnails/8.jpg)
*
*The projectile will experience two:
*Accelerations (ax= o and aY= -9.8m/s2)
*Velocities
*Displacements
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Upwardly Launched Projectiles
When a projectile is launched at an upward
angle, it follows a curved path and finally
hits the ground because of gravity.
The Vertical distance a cannonball falls below “imaginary
path if no gravity” is the same vertical distance it
would fall if it were dropped from rest & had been
falling for the same amount of time.
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*Draw a free body diagram with a coordinate
system.
*Divide the information into x and y components
*Look at your formulas and decided which
one(s) to use.
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Formulas for horizontal and vertical
motion of a projectile
Objects that have been thrown will have a
horizontal velocity that stays the same (no
horizontal acceleration ax = 0m/s2)
So vfx =vix in the second formula and third
formulas under horizontal motion.
(X) Horizontal (Y) Vertical
xf-xi = vixt + ½ axt 2 yf-yi = viyt + ½ ayt
2
vfx = vix + axt vfy = viy + ayt
vfx2 - vix
2 = 2ax(xf-xi) vfy2 = viy
2 + 2ay (yf-yi)
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*
This equation only works when y and y0 are both the same magnitude
𝑡 =2𝑦
𝑔
y0 y
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If a ball is thrown up in the air from a moving truck, where will it land?
(Ignore air resistance)
In front of the truck, behind the truck, or back in the truck
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Where will a package land if it is released
from a plane?
Behind the plane, in front of the plane
below the plane
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What is the horizontal distance covered by
an arrow that was shot through the air at a
600 angle with a velocity of 55 m/s?
Given
v = 55m/s
vx=27.5 m/s
vyo=47.6m/s
ax=0
ay=-9.8m/s2
t=?
x =?
Solve
Vx = cos 60(55m/s)=27.5 m/s
Vyo = sin60(55m/s)=47.6m/s
vy=vyo +ayt
x = Vx t (we need time)
600
55m/s y
Vx x
0 = 47.6m/s + -9.8m/s2 t
-47.6m/s = -9.8m/st
4.86 s =t
x = 27.5 m/s(9.7s)
x = 266.75m
Total time in the air 4.86s x 2 = 9.7s
Need to find time first!
To find x dist: x = vx t
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* A boat heading due north
crosses a river with a
speed of 10.0 km/h. The
water in the river has a
speed of 5.0 km/h due
east.
Moving frame of reference
In general we have PA PB BAv v v
(a) Determine the velocity of the boat.
(b) If the river is 3.0 km wide how long does it take to cross it?