In this lecture, for those with better mathematics, the following are introduced:
• More projectile motion examples .
• Accelerations that change with time.

Projectile motion examples

E2. A tennis ball is hit from a height of 2.45 m with an initial horizontal velocity component of 28 m.s^-1 and an unknown initial downward vertical velocity component.
The ball travels without deviating from its initial direction and hits the net 12 m away at a height of 0.5 m. Find
(a) the time of flight
(b) the initial vertical component of the velocity
(c) the angle from the horizontal that the ball was hit.

The assumption is that there are no drag forces due to air resistance, so the horizontal speed component stays constant and gravity acts on the vertical speed component.

(a)	Time of flight = (distance to net)/(horizontal speed component) = 12/28 = 4/7 s
(b)	The vertical distance fallen is 2.45 - 0.5 = 1.95m This is the distance travelled at the initial downward speed plus the extra distance due to gravitational acceleration. Height lost = 1.95 = uv×(4/7) + ½·9.8×(4/7)2 = (4/7)uv + 1.6 uv = (7/4)×(1.95 - 1.6) = 0.61 m.s-1 The initial vertical component of the velocity is downwards at 0.61 m.s-1
(c)	Tangent of angle from the horizontal = (initial vertical speed)/(constant horizontal speed) = 0.61/28 The angle from the horizontal = 1.250

E5. A golf ball is hit so that initially it has a velocity of 39.2 m.s^-1 at an angle of 30⁰ to the horizontal level fairway. The ball travels without deviating from its initial direction. Calculate:
(a) the time of flight, and
(b) the range of the golf ball along the fairway.

(a) From the displacement diagram: sin300 = (4.9·t02)/(39.2·t0) = t0/8 i.e. t0 = 8×sin300 = 4s The time of flight was 4s (b) From the displacement diagram: cos300 = S/(39.2×4) = S/156.8 i.e. S = 156.8×cos300 = 135.8m The range of the ball was close to 136m.

E9. At the Mexico City Olympics in 1968, Bob Beamon broke the long jump record with a jump of 8.90 m. Assume his initial take off speed was 9.5 m.s^-1 and he jumped at an angle of 37.4⁰ to the horizontal. Find
(a) the time he was in the air, and
(b) the accleration of gravity in Mexico City.

(a) cos37.40 = 8.9/(9.5×t0) i.e. t0 = 8.9/(9.5×cos37.40) = 1.18s The time of flight was 1.18s (b) sin37.40 = (½·g·t02)/(9.5×t0) = g·(1.18)/19 i.e. g = (19×sin37.40)/1.18 = 9.78m.s-2 The acceleration of gravity was 9.78m.s-2.

E4. A rugby football player positions a ball on the ground in the centre of the field and 40m from the goal posts. His kick gives the ball an initial speed of 20.8 m.s^-1 at an angle of 40⁰ to the horizontal. The ball travels without deviating towards the centre of the bar. The cross bar of the goal posts is 3 m above the ground. Find out if the ball clears the cross bar by calculating the height of the ball at the cross bar position.

cos400 = 40/(20.8×t0) t0 = 40/(20.8×cos400) = 2.5s The time of flight was 2.5s sin400 = [ 4.9×(2.5)2 + h ]/[ 20.8×2.5 ] i.e. sin400 = (30.6 + h)/52 Rearranging: h = 52×sin400 - 30.6 = 33.4 - 30.6 = 2.8m The ball was below the bar at 40m.

E6. A stone is thrown with an initial speed of 25 ms^-1 at a window in a vertical wall. The wall is 30 m away horizontally. The stone takes 2 s to hit the middle of the window. Assume no deviation from the initial direction and no air resistance. Calculate
(a) the angle (above the horizontal) at which the stone is thown, and
(b) how high the middle of the window is above the point where the stone is released.

(a) cosθ = 30/(25×2) = 0.6, i.e. θ = 53.10 The angle of elevation was 53.10 (b) sin53.10 = [ 4.9×(2)2 + h ]/(25×2) = (19.6 + h)/50 h = 50×sin53.10 - 19.6 = 40 - 19.6 = 20.4m The middle of the window was 20.4m above ground level.

E7. An athlete hurls a shotput 60m along a horizontal playing field. The shotput is released 2m above the ground and it takes 2.8 s to complete its trajectory and land on the ground. Find the angle above the horizontal at which the shotput was thrown and the initial speed of the shotput.

(a) tanθ = [ 4.9×t02 -2 ]/60          = [ 4.9×(2.8)2 - 2 ]/60 = 0.607     θ = 31.30 The angle of elevation was 31.30 (b) cos31.30 = 60/(2.8·v) v = 60/2.8×cos31.30 = 25m.s-1 The intitial velocity was 25m.s-1 at 31.30 up from the horizontal.

E8. A dart is thrown towards a dartboard. The dart is released at a height of 1.8 m, which is exactly the height of the centre of the bull’s eye of a dartboard. When it is released it is 3.04 m horizontally from the dartboard and moving upwards at an angle of 19⁰ to the horizontal. It hits the dartboard 180 mm directly below the bull’s eye. Find
(a) the time of flight of the dart,
(b) the initial speed of the dart,
(c) the final speed of the dart as it just hits the dartboard.

(a) tan190 = [ 4.9×t02 - 0.18 ]/3.04         t02 = [ 3.04×tan190 + 0.18 ]/4.9 = 0.25 t0 = 0.5s. The time of flight was 0.5s. (b) cos190 = 3.04/(0.5·v) v = 3.04/0.5×cos190 = 6.43m.s-1 The intitial speed was 6.43m.s-1.	(c) The angle at the top of the triangle is 710. The vertical side is 9.8×0.5 = 4.9m.s-1. With two sides and the included angle, the cos rule can be used to find the third side. Vfinal2 = 6.432 + 4.92 - 2×6.43×4.9×cos710 Vfinal2 = 44.8 Vfinal = 6.7m.s-1

Notice that the angles for the resultant displacement and the final velocity are different.
For the displacement: α = tan^-1(0.18/3.04) = 3.4⁰
For the velocity (from the sine rule): β = sin^-1(4.9×sin71⁰/6.7) - 19 = 24.75⁰.

This is a harder question as you need to be able to solve quadratic equations.

E3. A rescue plane drops urgently needed supplies to some scientists stranded on a flat, level iceberg. When the plane is 450 m North of the scientists and 310 m higher than them the pilot releases the supplies. When the supplies leave the plane the plane and supplies are travelling South at 52.6ms^-1 and ascending at an angle of 20⁰ to the horizontal. The supplies fall with normal gravitational acceleration and land on the iceberg. Find
(a) the time for the supplies to hit the ground.
(b) where the supplies land relative to the scientists.

From the displacement diagram top triangle: sin200 = (4.9·t02 - 310)/(52.6·t0) Rearranging: 4.9·t02 - 52.6·sin200·t0 - 310 = 0 or: 4.9·t02 - 18·t0 - 310 = 0 Using the theory of quadratic equations: t0 = [ -(-18) ± √((-18)2 - 4×(+4.9)×(-310)) ]/(2×4.9)     = [ 18 ± √(324 + 6076) ]/9.8 i.e. t0 = (18 ± 80)/9.8 Which gives: t0 = +10s or -6.3s As -6.3s was before the supplies were dropped, this can be discarded as imaginary. Hence time of flight is 10s.

R is the horizontal distance travelled in 10s.
From the displacement diagram top triangle: R = 52.6×10×cos20⁰ = 494m
This is 450 - 494 = -44m North, i.e. 44m South of the scientists. (They probably ducked).

Variable Acceleration

Don't worry too much about this if you are brand new to Physics. The things to look for are patterns.

In the following example graphs, these concepts are used:
• the area under the acceleration vs time graph gives the change in speed, and
• the area under the speed vs time graph gives the change in position.

First column: A zero acceleration (zero area) gives a constant speed (rectangle). A constant speed (rectangle) gives a position that increases linearly with time (triangle). Second column: A non-zero constant acceleration (rectangle) gives a speed that increases linearly with time (triangle). A speed that increases with time (triangle) gives a position that increases with half the square of time (curve). Third column: An acceleration that increases with time (triangle) gives a speed that increases with half the square of time (curve). A speed that increases with the square of time (curve) gives a position that increases with …

This is how the time variable changes as you move up and down between graphs.
To move from acceleration, to speed, to position, the pattern is to raise the time index by one and divide by the increased index.
To move from position, to speed, to acceleration, the pattern is to multiply by the time index and to reduce the index by one.

Examples

12. The deceleration of a boat along the x-axis is given by a = - 6t m.s^-2, where t > 0 s and t < 2.2 s. The boat is +3 m from the origin of the x-axis and travelling with a speed of +12 m.s^-1 when the time t = 0 s. Find
(a) the time taken for the boat to (momentarily) come to rest.
(b) the displacement when the boat is at rest.

When the speed is zero: At 2 s the boat is at rest. At 2 s the position is: When the boat is at rest it is +19 m from the origin.

Tutorial Questions

13. A rocket driven sled running on a straight track gives an acceleration of (2.4x10^-2)t m.s^-2, where t is the time from when it started. Find
(a) the speed it reaches after 100 s of this acceleration.
(b) the distance it travels in that 100 s.

14. The speed of an object along the x-axis is given by v = 0.5t2 + 3 m.s^-1, where t "e 0 s and t < 10 s. When the timing starts, the object is at the origin of the x-axis. Find
(a) the acceleration at 6 s.
(b) the displacement at 6 s.

15. The displacement of a particle from the x axis origin is x = (3t^-2 - t³) m, where t > 0 s. Find
(a) the time when it reaches its maximum positive x value.
(b) the maximum positive x value.
(c) the x value when its acceleration is zero.

16. At time t = 0 s, a traffic light turns green, and a car starts off from the lights with an acceleration of 0.6t m.s^-2. Also at time t = 0 s, a truck travelling in the same direction at a constant speed of 10 m.s^-1 passes the lights (and the car). Find
(a) the distance from the lights that the car catches up to the truck.
(b) the speed of the car when it reaches the truck.

Summarising:

To move from acceleration, to speed, to position, the pattern is to raise the time index by one and divide by the increased index.
To move from position, to speed, to acceleration, the pattern is to multiply by the time index and to reduce the index by one.

Acknowledgement: I learned this way of describing projectile motion from Dr David Wheeler (Thanks Dave!).

 Peter's Index   Physics Home   Lecture 5   Top of Page   Lecture 7 

email Write me a note if you found this useful