Topic&2:&Mechanics& 2.1$–Mo(on$ndscienceclasses.weebly.com/uploads/1/1/1/0/111005487/projectilemotion... · Analysing projectile motion Topic&2:&Mechanics& 2.1$–Mo(on$ •The

Projectile motion

•A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity. •Baseballs, stones, or bullets are all examples of projectiles executing projectile motion. •You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car).

Topic 2: Mechanics 2.1 – Mo(on

FYI •We will ignore air resistance in the discussion that follows…

Analysing projectile motion

•Regardless of the air resistance, the vertical and the horizontal components of velocity of an object in projectile motion are independent.


Slo

win

g do

wn

in +

y di

r. Speeding up in -y dir.

Constant speed in +x dir. ax = 0

a y =

-g a

y = -g

Analysing projectile motion •The trajectory of a projectile in the absence of air is parabolic. Know this!


Analysing projectile motion with fluid resistance •If there is air resistance, it is proportional to the square of the velocity. Thus, when the ball moves fast its deceleration is greater than when it moves slow.


SKETCH POINTS

•Peak to left of original one.

•Pre-peak distance more than post-peak.



PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (e) Sketch the following graphs: a vs. t, vx vs. t, vy vs. t: SOLUTION: The only acceleration is g in the –y-direction. •vx = 54, a constant. Thus it does not change over time. •vy = 15 - 10t Thus it is linear with a negative gradient and it crosses the time axis at 1.5 s.

t ay

-10

t

vx 54

t

vy 15 1.5

tailored equations for this particular projectile

∆x = 54t vx = 54

∆y = 15t - 5t 2

vy = 15 - 10t



• Answer next page



•The acceleration is ALWAYS g for projectile motion-since it is caused by Earth and its field. •At the maximum height the projectile switches from upward to downward motion. vy = 0 at switch.






•The flight time is limited by the y motion.

•The maximum height is limited by the y motion.






•The horizontal component of velocity is vx = ux which is CONSTANT. •The vertical component of velocity is vy = uy – 10t , which is INCREASING (negatively).





0.0s

0.5s

4 m

ux = ∆x / ∆t = (4 - 0) / (0.5 - 0.0) = 8 ms-1.





0.0s

0.5s

11 m

sy = uyt + 0.5ayt2



0.0s

0.5s

11 m

11 = uy(0.5) + 0.5(-10)(0.5)2

11 m [UP]

-10 ms-2 [UP]

Respect the vector Be consistent



0.0s

0.5s

11 m

11 = uy(0.5) - 1.25



0.0s

0.5s

11 m

11+1.25 = uy(0.5)



0.0s

0.5s

11 m

12.25= uy(0.5)



0.0s

0.5s

11 m

24.5= uy





•New peak below and left.

•Pre-peak greater than post-peak.

Source Material provided by Timothy K Lund

Adapted by J Sciarre?a for Extended Physics

Topic&2:&Mechanics& 2.1$–Mo(on$ndscienceclasses.weebly.com/uploads/1/1/1/0/111005487/projectilemotion... · Analysing projectile motion Topic&2:&Mechanics& 2.1$–Mo(on$ •The

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