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Peter's Index Physics Home Lecture 4 Course Index Lecture 6
Bridging Course  Lecture 5 (Projectile Motion)
In this lecture the "Wheeler" method of treating projectile motion is introduced.
Traditionally, projectile motion has been done by taking components of the displacements and velocities.
However this loses sight of the essential vector nature of projectile motion.
Dave Wheeler showed me this way of treating projectile motion while I was at UNSW and he was in Thailand.
It homes in on the end points of the motion, without needing to know the details of the actual trajectory.
Whole vectors give the benefits of simplicity and insight into the differing effects of initial velocity and the acceleration.
The really interesting motion of projectiles includes:
• the viscous drag of the air it passes though,
• air density variations (buoyancy),
• the change of gravitational acceleration with height, and
• the shape and rotation of the object (Golf ball dimples and
Magnus force).
The trajectory of a real projectile through the air, therefore, has to be looked up in empirical tables, or calculated with computers by iteration, that is, by calculating the change from one point to an adjacent point and repeating this indefinitely ….
This course doesn't cover all this really interesting material. As an introduction, the simplest case is treated, where the effect of viscous drag, density and gravity variations etc are neglected. The only acceleration on the projectile (once it is released from its projecting mechanism) is taken to be the constant vertical acceleration of gravity.
Magnitude graphs In terms of vertical motion only: an object with an initial upward speed, +v_{1}, from initial height, +h_{0}, under a downwards constant gravitational acceleration gives the following graphs. Note that Δx_{g} is the position change due to acceleration, and is the area of the triangle bounded by +v_{1}, t, and (t,v_{1}gt). 
Velocity vector diagram
The vector equation for initial and final velocities under gravity is
v_{2} = v_{1} + gΔt Displacement vector diagram Including magnitude and direction, the general vector diagram for displacements is: Where v_{1}Δt is the final displacement that would occur without the acceleration, and ½gΔt^{2} is the extra displacement due to the acceleration. 
Projectile Motion Examples
25. A shell is fired with a muzzle velocity of 500 m.s^{1} from a gun on a horizontal plane.
The barrel of the gun points upwards from the horizontal at an angle of 36°. Find
(a) the maximum height of the shell's trajectory.
(b) the range of the shell's trajectory.

At maximum height, the shell is no longer moving vertically and is (momentarily) just moving horizontally,
so the velocity diagram is: 
From this it can be seen that the time, t_{0}, to get to maximum height is given by: 
The displacement diagram for maximum height is 
Knowing the time to maximum height, t_{0}, the maximum height can be calculated from: 
The range is the horizontal distance to where the shell falls back level.
The displacement diagram is: 
From this, the time of flight, t_{1}, can be calculated: Knowing the time of flight, t_{1}, the range can be calculated from the same diagram: 
26. A cricket ball leaves the bat with a speed of 40 m.s^{1} at an angle of 60° upwards from the horizontal. If it is not caught, find how far away it lands.

The displacement diagram is 
The time of flight, t_{1}, is given by Knowing the time of flight, t_{1}, the range, Δs, is given by 
27. A Kangaroo can jump 8.1 m. Find its takeoff speed if it jumps at an angle of 45°.

The displacement diagram is 
The time of flight, t_{1}, is given by Knowing the time of flight, t_{1}, the take off speed, v_{0}, is given by 
28. An object is hurled from the top of a vertical cliff 196 m high.
The object is projected downwards at an angle of 29.4° to the horizontal with an initial speed of 30 m.s^{1}. Find
(a) the time it takes to hit the ground.
(b) how far away from the foot of the cliff it lands.

The displacement diagram is 
The time of flight is t_{0}. 
The horizontal range can be calculated from the right angled triangle at the top, by:.
Range = 30t_{0}cos29.4^{0} = 30×5×cos29.4^{0} = 131 m.
29. A tennis ball is served horizontally at a height of 2.5 m and just clears the net which is 12 m away and 0.9 m high. Find
(a) the minimum initial velocity.
(b) where it would land given the initial velocity of part (a).

The time to the net is t_{0} and the displacement diagram is: 
Comparing the heights on either side, 
The time of flight is the time to fall all the 2.5m to the ground under gravity, which gives:
2.5 = 4.9t_{1}^{2}
i.e. t_{1}^{2} = 2.5/4.9
giving t_{1} = 0.71s.
Moving horizontally at 21m.s^{1}, in 0.71s the ball will have travelled 21 × 0.71s = 14.9m
30. A hot air balloon is drifting steadily Eastwards at 5 m.s^{1} and
ascending at 9.8 m.s^{1} from a plain. A sandbag is released when the balloon is at 308.7 m. Find
(a) the maximum height that the sandbag reaches.
(b) the displacement of the sandbag from its point of release to where it hits the ground.

A velocity of 5m.s^{1} East and 9.8m.s^{1} up, gives the vector v_{1}
as shown on the left hand diagram. 

The displacement, Δs, for maximum height is the sum of the displacements due to initial velocities
and the effect of acceleration, with t_{0} = 1s. 

t_{1} is the time of flight. 
In Cartesian coordinates, the displacement from the point of release is
(5t_{1} , 308.7), i.e. (45 , 308.7)m
You should check that you can convert this to Polar coordinates to get (312m , 81.7^{0})
31. A motorcycle stunt rider leaves a ramp inclined at 30^{°} to the horizontal at a height of 6.125 m
and at a speed of 19.6 m s^{1}. He jumps over an obstacle and lands on the ground. Find
(a) the time for maximum height.
(b) the maximum height above ground
(c) the time of flight.
(d) the horizontal distance travelled in the air.

The time for maximum height, t_{0}, is given by a velocity diagram for maximum height. 

The maximum height is given by a displacement diagram, where the blue arrow shown is the displacement for maximum height. 

The time of flight, t_{1}, is given by a displacement diagram for the end point of the motion. 
The horizontal distance travelled in the air is given by cos30^{0} = Range/19.6t_{1}
i.e. The Range = 19.6×2.5×cos30^{0} = 42.4 m
Summarising:
The general velocity diagram for projectiles is: 
The general displacement diagram for projectiles is: 
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