Projecles. ! Mo(on!of!Objects!Projected! Horizontally… 03... · Height: maximum vertical distance ... A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff

 Projec(les.  Mo(on  of  Objects  Projected  Horizontally  and  Ver(cally  

y

0

Lets  Review  a  free  falling  par(cle(  no  horizontal  mo(on):  

Time(s)   Vinst(m/s)   Δx(m)   Accel  (m/s2)  

1  sec   -­‐9.8  m/s2  

2  sec    

-­‐9.8  m/s2    

3  sec   -­‐9.8  m/s2    

4  sec   -­‐9.8  m/s2    

How would you Calculate the v and y of the particle?

ANALYSIS OF Projectile Motion ASSUMPTIONS:

•  x-direction (horizontal) uniform motion

•  y-direction (vertical) accelerated motion

•  no air resistance (need calculus..gets messy…)

QUESTIONS:

•  What is the trajectory?

•  What is the total time of the motion?

•  What is the horizontal range?

•  What is the final velocity?

Projec(le  Mo(on  

x

y

€

Horizontal (x)x = vxtVertical (y)

y = v0yt −12gt 2

vy = v0y − gt

v0

x

y

x

y

• Motion is accelerated

• Acceleration is constant, and downward

•  a = g = -9.81m/s2

• The horizontal (x) component of velocity is constant

• The horizontal and vertical motions are independent of each other, but they have a common time

g = -9.81m/s2

Part  2.  Mo(on  of  objects  projected  at  

an  angle  

#1 # 2 # 3 3 • All of these projectiles are considered free particle types

• That means no other forces act on them , just gravity (Fg)

• Lets graph the vertical and horizontal components of #3

#1 #1 #2 #3

velocity position

X

Y

acceleration

x

y

•  Motion is accelerated

•  Acceleration is constant, and downward

•  a = g = -9.81m/s2

•  The horizontal (x) component of velocity is constant

•  The horizontal and vertical motions are independent of each other, but they have a common time

Trajectory  and  horizontal  range  2

22 cos2tan x

vgxy

i Θ+Θ=

0

5

10

15

20

25

30

35

0 20 40 60 80

15 deg30 deg45 deg60 deg75 deg

vi = 25 m/s

WARNING! The following 3 slides

only apply to projectiles that have the same

launching and landing heights!

 Special  equa+ons  for  range  and  height!!!    

Range Height

Total Time in Air

€

R =vi2Sin2θg

€

h =vi2Sin2θ2g

€

T =2vyg

Range: horizontal distance traveled by a projectile

€

R =vi2 sin2θ

g

R θ

Double the angle first, then take the sine of that product! (for example )

€

60°

€

sin(120°) = 0.866

Height: maximum vertical distance of a projectile from launch point

€

h =vi2 sin2θ2g

h θ

Take the sine’s fraction (for example ) and square it!

€

60°

€

(.866)2 = 0.7450

Total Time: To find the total time, we need the y component of velocity

h θ

Take the y component and double it!

€

T =2vyg

€

20 m / s

€

θ

vy

A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high.

a) How long does it take for the stone to reach the bottom of the cliff?

b) How far from the base of the cliff does the stone hit the ground?

c) What are the horizontal and vertical components of the stone’s velocity just before it hits the ground?

A football is kicked with a velocity of 23 m/s at a 50° angle.

a) how far does it travel?

b) what is its maximum height?

c) how long is it in the air?

A ball is projected horizontally with a v0 of +8 m/s. Find its position and velocity after 0.25 seconds after the launch.

Core Problem 2.0 A ball is projected upward with an initial velocity of 160 m/s directed at an angle of 53 degrees to the horizon.

a) Find the v0x and v0y of the ball at t= 0 b) Find the position and magnitude of the ball when t = 2 secs c) What is the maximum height reached of this projectile. d) What is the ball’s range?