Topic 9: Motion in fields 9.1 Projectile motion

9.1.1 State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field.

9.1.2 Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.

9.1.3 Describe qualitatively the effect of air resistance on the trajectory of a projectile.

9.1.4 Solve problems on projectile motion.

Topic 9: Motion in fields9.1 Projectile motion

State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field.

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.

You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car).


FYIWe will ignore air resistance in the discussion that follows…

State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field.

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


Slowing down in +y

dir.

Speeding up in -y

dir.

Constant speed in +x dir. ax = 0

ay = -g

ay = -g

Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.

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


Describe qualitatively the effect of air resistance on the trajectory of a projectile.

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 POINTSPeak to left of

original one.

Pre-peak distance more than post-peak.

Solve problems on projectile motion.

Recall the kinematic equations from Topic 2:

Since we worked only in 1D at the time, we didn’t have to distinguish between x and y in these equations.

Now we appropriately modify the above to meet our new requirements of simultaneous equations:


kinematic equations

s = ut + (1/2)at2

v = u + at a is constant

Displacement

Velocity

kinematic equations

∆x = uxt + (1/2)axt2

vx = ux + axtax and ay are constant∆y = uyt + (1/2)ayt2

vy = uy + ayt



kinematic equations

∆x = uxt + (1/2)axt2

vx = ux + axtax and ay are constant∆y = uyt + (1/2)ayt2

vy = uy + ayt

PRACTICE: Show that the reduced equations for projectile motion are

SOLUTION: ax = 0 in the absence of air resistance.ay = -10 in the absence of air resistance.

0

0

reduced equations of projectile

motion

∆x = uxt

vx = ux

∆y = uyt - 5t2

vy = uy - 10t



EXAMPLE: Use the reduced equations above to prove that projectile motion (in the absence of air resistance) is parabolic.

SOLUTION: Just solve for t in the first equation and substitute it into the second equation.

∆x = uxt becomes t = ∆x/ux so that t2 = ∆x2/ux2.

Then ∆y = uyt - 5t2, or

∆y = (uy/ux)∆x – (5/ux2)∆x2.

FYIThe equation of a parabola is y = Ax + Bx2.In this case, A = uy/ux and B = -5/ux

2.


motion

∆x = uxt

vx = ux

∆y = uyt - 5t2

vy = uy - 10t




motion

∆x = uxt

vx = ux

∆y = uyt - 5t2

vy = uy - 10t

PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (a) What are ux and uy?SOLUTION: Make a velocity triangle. u = 5

6 m s-1

ux = u cos

uy = u sin = 15º

ux = 56 cos 15ºux = 54 m s-1

uy = 56 sin 15ºuy = 15 m s-1.




motion

∆x = uxt

vx = ux

∆y = uyt - 5t2

vy = uy - 10t

PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (b) What are the tailored equations of motion? (c) When will the ball reach its maximum height?SOLUTION: (b) Just substitute ux = 54 and uy = 15:

(c) At the maximum height, vy = 0. Why? Thusvy = 15 - 10t becomes 0 = 15 - 10t so that 10t = 15 t = 1.5 s.

tailored equations for this particular

projectile

∆x = 54t

vx = 54

∆y = 15t - 5t2

vy = 15 - 10t




motion

∆x = uxt

vx = ux

∆y = uyt - 5t2

vy = uy - 10t

PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (d) How far from the muzzle will the ball be when it reaches the height of the muzzle at the end of its trajectory?SOLUTION: From symmetry tup = tdown = 1.5 s so t = 3.0 s. Thus ∆x = 54t ∆x = 54(3.0) ∆x = 160 m.




motion

∆x = uxt

vx = ux

∆y = uyt - 5t2

vy = uy - 10t

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, and 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.

tay

-10

tvx

54

tvy15

1.5



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.



ax = 0.

ay = -10 ms-2.



Fall time limited by y-equations:

∆y = uyt - 5t2

-33 = 0t - 5t2

-33 = -5t2

(33/5) = t2

t = 2.6 s.



Use x-equations and t = 2.6 s:

∆x = uxt

∆x = 18(2.6)

∆x = 15 m.



vx = ux

vy = uy – 10t

vx = 18.

vy = 0 – 10t

vy = –10(2.6) = -26.

18

26

tan = 26/18

= tan-1(26/18) = 55º.



The horizontal component of velocity is vx = ux

which is CONSTANT.

The vertical component of velocity is vy = uy – 10t which is INCREASING (negatively).



∆EK + ∆EP = 0

∆EK = -∆EP

∆EK = -mg∆y

∆EK = -(0.44)(9.8)(-32) = +138 J = EK – EK0

EK = +138 + (1/2)(0.44)(222) = 240 J.

EK0 = (1/2)mu2



If 34% of the energy is consumed, 76% remains. 0.76(240) = 180 J

(1/2)(0.44)v2 = 180 J

v = 29 ms-1.



Use ∆EK + ∆EP = 0.

(1/2)mvf2 - (1/2)mv2 = -∆EP

mvf2 = mv2 + -2mg(0-H)

vf2 = v2 + 2gH



ux = u cos ux = 28 cos 30º

ux = 24 m s-1.

uy = u sin uy = 28 sin 30º

uy = 14 m s-1.



∆x = uxt16 = 24t

t = 16/24 = 0.67

∆y = uyt – 5t2

∆y = 14t – 5t2

∆y = 14(0.67) – 5(0.67)2 = 7.1 m.

The time to the wall is found from ∆x…



0.0s

0.5s

4 m

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



0.0s

0.5s

4 m

11 m

uy = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms-1.



0.0s

0.5s

4 m

11 m

1.0s

1.5s2.0s

2.5s 3.0s

24 m

30 m

D2 = 242 + 302 so that D = 38 m

D

= tan-1(30/24) = 51º

,@ = 51º.



New peak below and left.

Pre-peak greater than post-peak.

Topic 9: Motion in fields 9.1 Projectile motion

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