Problems will be solvable using the equations of uniform acceleration Projectile motion This section is devoted to understanding the motion of projectiles A projectile is an object upon which the only force acting is gravity ID: 927744
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Slide1
Specification
Independent effect of motion in horizontal and vertical directions of a uniform gravitational field.
Problems will be solvable using the equations of uniform acceleration.
Projectile motion
Slide2This section is devoted to understanding the motion of projectiles
A projectile is an object upon which the only force acting is gravity
An object dropped from rest (or thrown down with an initial velocity) is a projectile (provided that the influence of air resistance is negligible)
An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible)And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible)
Projectile motion
Slide3We only need to consider motion in 1-Direction: vertically
If it is dropped from rest then the initial velocity,
u = 0 ms
-1An object dropped from rest (or thrown down with an initial velocity) is a projectile (provided that the influence of air resistance is negligible)The motion is in the direction of the acceleration: a = g = +9.81 ms-2
Projectile motion
Slide4An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible)
We only need to consider motion in 1-Direction: vertically (up and down)
At the maximum height,
smax, the velocity, v = 0 ms-1The motion is initially in the opposite direction of the acceleration: a = g = -9.81 ms-2
Projectile motion
Slide5We only need to consider motion in 1-Direction: vertically (up and down and past the ground level)
At the maximum height,
s
max, the velocity, v = 0 ms-1The velocity,
v
, when the ball has reached the initial position,
s = 0
is the same as the initial velocity,
u
An object that is thrown vertically upward is also a projectile and drops beyond the point it was released (provided that the influence of air resistance is negligible)
The motion is initially in the opposite direction of the acceleration:
a = g = -9.81 ms
-2
Projectile motion
Slide6Problem:
A man stands on the edge of the cliff and throws a stone vertically upwards at 15ms
-1
. After what time will the stone hit the ground 20m below?a = g = -9.81 ms-2
At the maximum height,
s
max
, the velocity,
v = 0 ms
-1
The answer is ~4s, you work it out!
Option 1:
Calculate the time and distance for the journey to the top,
Assume it is being dropped from the top at rest with a new value for s: distance to the top + 20m, and calculate the time
Add both times together
Option 2:
Calculate the time to travel all the way up and back to its original height, so s = 0m
Calculate the time from ground level to 20m assuming it was dropped with initial speed the same as the original speed
Projectile motion
Slide7A
B
A golfer hit ball A. It took 5 seconds to land in the position shown.
He then hit ball B with less force. How long did it take to land in its position?
4 seconds
5 seconds
6 seconds
Projectile motion
Slide8Projectiles in 2 directions
Horizontally and vertically
Projectile motion
Slide9Package drop
The package follows a parabolic path and remains directly below the plane at all timesThe vertical velocity changes (faster, faster)
The horizontal velocity is constant!
Slide10The vertical velocity changes (faster, faster)
The horizontal velocity is constant!
Projectile motion
Slide11The time taken for the object to reach the ground along the parabolic path is the same as if it were dropped vertically.
The parabolic path for an object projected horizontally is shown in the diagram Notice that if air resistance is ignored the vertical height of the object at given times after the start is the same no matter what horizontal velocity it had at the moment of release.
Projectile motion
Slide12A parcel is released from an aircraft travelling horizontally at a speed of 120ms
-1
above level ground.
The parcel hits the ground 8.5 seconds later. Calculate:The height of the aircraft above the groundThe horizontal distance travelled in this time by i) the parcel, ii) the aircraftThe speed of impact of the parcel on the ground.
Taking vertical motion
s =
½
gt
2
s = ½ x 9.8 x 8.5
2
= 354 m
b. Taking horizontal motion s = v x t
s = 120 x 8.5 = 1020m
This is the distance travelled by the aircraft and the parcel
c. Taking vertical motion g = v/t
vertical speed v = 9.8 x 8.5 = 83.3 ms
-1
Horizontal speed = 120 ms
-1
(resultant speed)
2
= 83.3
2
+ 120
2
Resultant speed = 146 ms
-1
2. An object is projected horizontally at a speed of 16ms
-1
into the sea from a cliff top of height 45.0m. Calculate:
How long it takes to reach the seaHow far it travels horizontallyIts impact vertical velocity.Taking vertical motions=
½
gt
2
45 = ½ x 9.8 x t
2
t = 3 s
45.0m
b. Taking horizontal motion
s = v x t
s = 16 x 3 = 48m
c.
The vertical motion
g = v/t v =
gt
v = 9.8 x 3 = 29.4 ms
-1
16ms
-1
Slide143. A dart is thrown horizontally along a line passing through the centre of a dartboard, 2.3 m away from the point at which the dart was released. The dart hits the dartboard at a point 0.19m below the centre. Calculate:
The time of flight of the dart
Its horizontal speed of projection.
2.3 m
Taking vertical motion
s=
½
gt
2
0.19 = ½ x 9.8 x t
2
t = 0.197 s = 0.2 s
Taking horizontal motion
s = v x t v = s/t
v = 2.3/0.197 = 11.7 ms
-1
Slide15Physics Support MaterialsHigher
Mechanics and Properties of MatterA box is released from a plane travelling with a horizontal velocity of 300 m s
-1 and at a height of 300 m. Find:
43
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Equations of Motion
a) how long it takes the box to hit the ground
Consider the vertical motion of the box
Down is the positive direction
Slide16Physics Support MaterialsHigher
Mechanics and Properties of MatterA box is released from a plane travelling with a horizontal velocity of 300 m s
-1 and at a height of 300 m. Find:
43 continued
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Equations of Motion
b) the horizontal distance between the point of impact and the release point
Consider the horizontal motion of the box
c) the position of the plane relative to the box at the time of impact
The box and the plane have the same horizontal motion (constant speed), so they travel the distance horizontally as the box falls. At impact, the plane is vertically above the box.
Slide17Physics Support MaterialsHigher
Mechanics and Properties of MatterA projectile is fired horizontally from the edge of a cliff at 12 m s
-1 and hits the sea 60 m away. Find:
44
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Equations of Motion
the time of flight
the height of the starting point above the sea level
Consider the horizontal motion of the projectile
60 m
Sea
12 m s
-1
Physics Support MaterialsHigher
Mechanics and Properties of MatterA projectile is fired horizontally from the edge of a cliff at 12 m s
-1 and hits the sea 60 m away. Find:
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Equations of Motion
b) the height of the starting point above the sea level
60 m
Sea
12 m s
-1
Consider the vertical motion of the projectile
Down is the positive direction
Slide19Physics Support MaterialsHigher
Mechanics and Properties of MatterA ball is projected horizontally at 15 m s
-1 from the top of a vertical cliff. It reaches the horizontal ground 45 m from the foot of the cliff.
45
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Equations of Motion
a) Draw graphs, giving appropriate numerical values of the ball’s
i
) horizontal speed against time
5
10
0
1
3
v / m s
-1
t / s
15
Calculating the time of flight:
Consider the horizontal motion of the projectile
2
Slide20Physics Support MaterialsHigher
Mechanics and Properties of MatterA ball is projected horizontally at 15 m s
-1 from the top of a vertical cliff. It reaches the horizontal ground 45 m from the foot of the cliff.
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Equations of Motion
a) Draw graphs, giving appropriate numerical values of the ball’s
ii) vertical speed against time
10
20
0
1
3
v / m s
-1
t / s
30
Calculating the final vertical velocity:
Consider the vertical motion of the projectile
2
Slide21Physics Support MaterialsHigher
Mechanics and Properties of MatterA ball is projected horizontally at 15 m s
-1 from the top of a vertical cliff. It reaches the horizontal ground 45 m from the foot of the cliff.
45 continued
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Equations of Motion
b) Use a vector diagram to find the velocity of the ball 2 s after its projection.
After 2 s the horizontal velocity is 15 m s
-1
After 2 s the vertical velocity is 20 m s
-1
15
m s
-1
20
m s
-1
v
v
2
= 15
2
+ 20
2
v
2
= 625
v = 25
m s
-1