Transcript 7-2 Projectile Motion

7-2 Projectile Motion
Independence of Motion in 2-D
 Projectile is an object that has been given an intial
thrust (ignore air resistance)

Football, Bullet, Baseball
 Moves through the air under the force of gravity
 Path is called Trajectory
 If you know the force of the thrust, you can determine
trajectory
Independence of Motion in 2-D
 If you hit a golf ball,
throw a football, only
force acting on the
projectile is gravity (long
range)
 Gravity acts same in
vertical direction
 Horizontal force has no
effect on vertical
component
 Combination of vertical drop and horizontal moment
give object a parabola trajectory
Strategy
 Separate into vertical and horizontal motion problem
 Vertical motion is treated like a straight up or down
movement (g)
 Horizontal motion treated like a constant velocity
problem


No thrust and air drag neglected
No horizontal forces acting (a = 0)
 Motions are connected by time variable
 Solve for time in one of the dimensions and will give you other
Equations
 Y-direction: vy = -gt
 y = y0 – (1/2)gt2
 t = √ -2(y – y0)
g
 X-Direction
 x = x0 + vx0t
Stone is thrown horizontally at 15 m/s from top a
cliff 44 meters high
 A) How far from the base of the cliff does the stone
hit the ground?
 Known

X0 = 0
vx0 = 15 m/s
 Unknown
 X when y = -44 m
 V at that time
Fg = Fnet
y0 = 0
a
vy0 = 0 a = -g
How far from the base of the cliff does the stone hit the
ground?
 Y-direction
 y = y0 – (1/2)gt2



t = √ -2(y-y0)
so: = √ -2(y) / g
g
√ -2(-44 ) / -9.8 m/s2
= 3.0 sec
How far from the base of the cliff does the stone hit the
ground?
 X direction
 x = x0 + vx0t
 x = (15 m/s)(3.0 m/s)
 = 45 meters from the base
How fast is it moving the instant before it hits the ground?
 Vy = -gt
vx
 = -(9.80 m/s2)(3.0 s)
 = -29 m/s
 v =√ vx2 + vy2
 = √(15 m/s)2 + (-29 m/s)2
 = 33 m/s
vy
v
Projectiles Launched at an Angle
 Initial velocity has an initial horizontal and vertical
component

Rises with slowing speed and falls with gaining speed
 Max Height
 height of projectile when vertical velocity is zero and only has
horizontal component
 y = yi + Vyit – (1/2)gt2
 Range (R)
 Horizontal distance the projectile travels
Problem
 A ball is launched with an initial velocity of 4.47 m/s
at an angle of 66o above the horizontal.



A) What is the max height the object attained?
B) How long did it take the ball to return to the launching
height?
C) What was the range?
 Known
 xi = 0
yi = 0
 Unknown
 y when Vy = 0
vi = 4.47 m/s
t = ??
θ = 66o
x when y = 0
a = -g
Equations Needed
 Y direction:
 Vyi = vi sinθ
(4.47)sin 66o
 Vyi = 4.08 m/s
 *** Vy = Vyi – gt
 *** y = yi + Vyit – (1/2)gt2
 x direction
 Vxi = vi cos θ
 Vx = Vxi
 x = xi + vxit
What is the max height the object attained?
 Vy = 0, t = vyi / g
 t = (4.08 m/s) / (9.80 m/s2)
 t = 0.420 s
2
=
v
t
–
(1/2)gt
max
yi
 = (4.08 m/s)(0.420) – (1/2)(9.80)(0.420)2
 y max = 0.850 m
y
How long did it take the ball to return to the launching
height?
 y=0
 y = yi + Vyit – (1/2)gt2
 0 = 0 + Vyit – (1/2)gt2
 t = 2vyi / g
 = 2(4.08 m/s) / (9.80)
 t = 0.83
What was the range?
 x=R
 R = Vxit
 = (4.47 m/s)(cos 66o)(0.83 s)
 R = 1.5 m