Peter's Physics Pages

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, +v1, from initial height, +h0, under a downwards constant gravitational acceleration gives the following graphs. Note that Δxg is the position change due to acceleration, and is the area of the triangle bounded by +v1, t, and (t,v1-gt). Velocity vector diagram The vector equation for initial and final velocities under gravity is    v2 = v1 + gΔt Where the plus sign indicates vector addition and the direction of the acceleration is included in the vector symbol. Including magnitude and direction, the general vector diagram linking initial and final velocities is: Displacement vector diagram Including magnitude and direction, the general vector diagram for displacements is: Where v1Δt is the final displacement that would occur without the acceleration, and ½gΔt2 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.

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 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, t0, to get to maximum height is given by: The displacement diagram for maximum height is Knowing the time to maximum height, t0, 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, t1, can be calculated: Knowing the time of flight, t1, 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.

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 The displacement diagram is The time of flight, t1, is given by Knowing the time of flight, t1, 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°.

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 The displacement diagram is The time of flight, t1, is given by Knowing the time of flight, t1, the take off speed, v0, 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.

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 The displacement diagram is The time of flight is t0. From the right angled triangle at the top, h = 30t0sin29.40 The total height is the height change due to the initial velocity plus the height change due to the acceleration. That is: 196 = 30t0sin29.40 + 4.9t02. Simplifying and dividing by 4.9 gives: t02 + 3t0 - 40 = 0 Two numbers which multiply to be -40 and add to be +3, are 8 and -5, hence the equation becomes: (t0 - 5)(t0 + 8) = 0 This gives two solutions for t0, +5s and -8s. Discarding the negative time, as something that might have happened before the start of the question, the time of flight is given as t0 = 5s.

The horizontal range can be calculated from the right angled triangle at the top, by:.
Range = 30t0cos29.40 = 30×5×cos29.40 = 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).

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 The time to the net is t0 and the displacement diagram is: Comparing the heights on either side, 2.5 = 4.9t02 + 0.9, i.e. t02 = (2.5-0.9)/4.9 Discarding the negative solution, as something that might have happened before the start of the question, this gives t0 = 0.57s. From the top and bottom horizontal distances, v0 × 0.57 = 12, or v0 = 21m.s-1.

The time of flight is the time to fall all the 2.5m to the ground under gravity, which gives:
2.5 = 4.9t12     i.e. t12 = 2.5/4.9     giving t1 = 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 v1 as shown on the left hand diagram. At the highest point, the change in velocity due to the acceleration has brought the velocity down to the horizontal, as shown by v0 in the right hand diagram. By comparing diagrams, 9.8 = 9.8·t0, so the time to maximum height, t0, is 1s. The displacement, Δs, for maximum height is the sum of the displacements due to initial velocities and the effect of acceleration, with t0 = 1s. Equating vertical heights gives 308.7 + 9.8 = h + 4.9 i.e. h = 313.6m t1 is the time of flight. The displacement, Δr, from the point of release to where it hits the ground is the vector sum of the displacements due to initial velocities and the displacement which is due to acceleration, at time t1. Δs1 gives the effect of the displacements due to initial velocities and 4.9t12 gives the displacement which is due to acceleration. Equating vertical heights gives 4.9 t12 = 308.7 + 9.8t1 Re-arranging and dividing by 4.9: t12 - 2t1 - 68 = 0 There are two solutions: t1 = +9s or -7s. The negative solution is discarded, so t1 = +9s.

In Cartesian coordinates, the displacement from the point of release is (5t1 , -308.7), i.e. (45 , -308.7)m
You should check that you can convert this to Polar coordinates to get (312m , -81.70)

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, t0, is given by a velocity diagram for maximum height. From the right angled triangle: sin300 = 9.8t0/19.6 Simplifying gives t0 = 1s The maximum height is given by a displacement diagram, where the blue arrow shown is the displacement for maximum height. For the right angled triangle, the adjacent side to 300 has a length which is 4.9t02 plus (h - 6.125) From this: sin300 = (4.9t02 + h - 6.125)/19.6t0 With t0 = 1   19.6×sin300 = 4.9 + (h - 6.125) Simplifying: h = 9.8 - 4.9 + 6.125 = 11.025m The time of flight, t1, is given by a displacement diagram for the end point of the motion. For the right angled triangle, the adjacent side to 300 has a length which is 4.9t12 - 6.125 From this: sin300 = (4.9t12 - 6.125)/19.6t1 Simplifying: 9.8t1 = 4.9t12 - 6.125              or: t12 - 2t1 - 1.25 = 0 Two differently signed numbers that add to be -2 and multiply to be -1.25, are -2.5 and +0.5 The equation can be written as (t1 - 2.5)(t1 + 0.5) = 0 i.e. the time of flight is 2.5s.

The horizontal distance travelled in the air is given by cos300 = Range/19.6t1
i.e. The Range = 19.6×2.5×cos300 = 42.4 m

Summarising:

 The general velocity diagram for projectiles is: The general displacement diagram for projectiles is:  email Write me a note if you found this useful

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