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Projectile Motion Projectile Motion

Projectile Motion - PowerPoint Presentation

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Uploaded On 2016-11-01

Projectile Motion - PPT Presentation

Objects Projected Horizontally Horizontal projectile motion   horizontally the velocity is always constant 0 vertically the initial velocity y is always zero vertically ID: 483376

velocity horizontal horizontally object horizontal velocity object horizontally ball 71392 time height vertical initial 8077 constant ertical speed projectile final airbornethe calculate

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Slide1

Projectile MotionObjects Projected Horizontally

Horizontal projectile motion:

 horizontally the velocity () is always constant ( = 0)vertically the initial velocity (y) is always zerovertically the object experiences downward acceleration due to gravity (= 9.81 m/s2)the horizontal and vertical motion is independent of each other except they have a common timethe time it takes for the object to hit the ground is not affected by the horizontal velocity  

 Slide2

Horizontal (x)

Vertical (y)

(range) (height) (constant) = 0 (constant)

= 0

= 9.81 m/s 2t (same)t (same)

Horizontal (x)Vertical (y)t (same)t (same)

animationSlide3

An object is kicked off a cliff with a horizontal velocity of 4.2 m/s. The height of the cliff is 3.2 m. Calculate the following:

the time the object is airbornethe horizontal displacement when it landsthe velocity upon entering the water[down] = +

ve[right] = +vevertical (y)horizontal (x) = 0 = 9.81 m/s2 = 3.2 mt = ? = ?

 

= 4.2 m/s = 0

= ?t = ? = 4.2 m/s Slide4

=

t +

3.2 = (0)t +0.5(9.81)4.905 = 3.2t = 0.8077s  = 4.2 m/s

t = 0.8077 s

= ? 

= t = (4.2)(0.8077) = 3.39 m  (a)(b)

=

+ t

= 0 + (9.81)(0.8077)

= 7.924 [down]

 

(c)

= 4.2 m/s

 

=

7.924

m/s

 

 

=

= 8.9

 

tan x =

x = 62°

 

\

The final velocity is 8.9 m/s [62° below the horizontal]

xSlide5

A baseball pitcher throws a ball horizontally. The ball falls 83 cm in travelling 18.4 m to the plate. Determine the ball’s initial horizontal speed.

[down] = +ve[right] = +ve

vertical (y)horizontal (x) = 0 = 9.81 m/s2 = .83 mt = ? 

=

? = 0 = 18.4 mt = ?

  = t + .83 = (0)t + 0.5(9.81) t = 0.411 s

 

=

=

= 44.77

 

\

Initial horizontal speed was 45 m/s [horizontal]Slide6

A tennis player serves a ball horizontally, giving it a speed of 24 m/s [horizontal] from a height of 2.5 m. Calculate the following

:the time the ball is airbornethe horizontal displacementthe velocity at impact[down] = +

ve[right] = +vevertical (y)horizontal (x) = 0 = 9.81 m/s2 = 2.5 mt = ? 

= 24 m/s

= 0 = ?

t = ?  = t + 2.5 = (0)t + 0.5(9.81)

t = 0.71392 s 

=

t

= (24)(0.71392)

= 17.13

 

\

horizontal displacement was 17 m[horizontal]Slide7

=

+

t = 0 + (9.81)(0.71392) = 7.004 [down]   = 24 m/s

 

= 7.004 m/s

   =

= 25

 

tan x =

x = 16°

 

\

The final velocity is 25 m/s [16° below the horizontal]

x

= 0

= 9.81

t = 0.71392

= ?