9.1.1state the independence of the vertical and the horizontal components of velocity for a...
TRANSCRIPT
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).
Topic 9: Motion in fields9.1 Projectile motion
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.
Topic 9: Motion in fields9.1 Projectile motion
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!
Topic 9: Motion in fields9.1 Projectile motion
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.
Topic 9: Motion in fields9.1 Projectile motion
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:
Topic 9: Motion in fields9.1 Projectile motion
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
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
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
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
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/ux
2.
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.
reduced equations of projectile
motion
∆x = uxt
vx = ux
∆y = uyt - 5t2
vy = uy - 10t
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
reduced equations of projectile
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.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
reduced equations of projectile
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
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
reduced equations of projectile
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.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
reduced equations of projectile
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
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
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.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
The flight time is limited by the y motion.The maximum height is limited by the y motion.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
ax = 0.
ay = -10 ms-2.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
Fall time limited by y-equations:
∆y = uyt - 5t2
-33 = 0t - 5t2
-33 = -5t2
(33/5) = t2
t = 2.6 s.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
Use x-equations and t = 2.6 s:
∆x = uxt
∆x = 18(2.6)
∆x = 15 m.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
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.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile 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).
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
∆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
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
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.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
Use ∆EK + ∆EP = 0.
(1/2)mvf2 - (1/2)mv2 = -∆EP
mvf2 = mv2 + -2mg(0-H)
vf2 = v2 + 2gH
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
ux = u cos ux = 28 cos 30º
ux = 24 m s-1.
uy = u sin uy = 28 sin 30º
uy = 14 m s-1.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
∆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…
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
0.0s
0.5s
4 m
ux = ∆x/∆t = (4-0)/(0.5-0.0) = 8 ms-1.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
0.0s
0.5s
4 m
11 m
uy = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms-1.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
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.
Solve problems on projectile motion.
Topic 9: Motion in fields9.1 Projectile motion
New peak below and left.
Pre-peak greater than post-peak.