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Question 3: Projectiles
Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier
- Keys for answering questions Page
- Projectiles keys: Test
- Ordinary Level Exam Questions: Particle fired horizontally
- Initial velocity is given in terms of i and j components
- Initial velocity needs to be resolved into i and j components
- Answers to Ordinary Level Exam Questions
- Higher Level: Particle fired from a horizontal surface: Target Practice
- Higher Level: Particle fired from a horizontal surface: General Questions
- Particle fired from an inclined plane: Strikes the plane at right angles
- Particle is moving horizontally when it strikes the inclined plane
- Particle fired from an inclined plane – general questions
- Maximum Range
- Combining Collisions with Projectiles
- Guide to answering Exam Questions 2010 -
Keys for answering questions
- How would you find time of flight? Answer: Find t when sy = 0.
- How would you find the range of a particle? Answer: Find t when sy = 0 and sub this time in to sx.
- How would you find the Maximum height reached by a particle? Answer: At maximum height vy = 0, so find t from this and sub this value in to sy. Or Use the equation v^2 = u^2 + 2as for the y direction (which becomes 0 = uy^2 + 2(-9.8)s)
- How would you find the magnitude of a particle’s velocity after 3 seconds? Answer: First find vx and vy when t = 3. Magnitude is then found from the formula Magnitude = √ (vx^2 + vy^2 )
- How would you find the direction of a particle after 3 seconds? Answer: First find vx and vy when t = 3. Direction is then found from the formula Direction = Tan-1^ (vy / vx)
- If an object is fired from a cliff which is 200 metres above sea-level, how will this affect our approach to answering the question? Answer: sy = - 200
- If a bullet is fired horizontally what does it mean? Answer: uy = 0
- When would we ever take ‘g’ to be positive? Answer: When an object is given an initial velocity downwards (this rarely comes up).
- If two projectiles collide in mid-air then.. .? Answer: (i) sy for first projectile = sy for second projectile (ii) sx for first projectile plus sx for second projectile equals the total distance between them at the beginning.
- If a projectile just clears a wall (or hits a dart-board) which is 3 m away and 5 m high then.. .? Answer: Sx = 3 when Sy = 5
Projectiles keys: Test
Note that time to reach maximum height is half of the time for full flight (if starting and finishing on horizontal ground).
- How would you find time of flight?
- How would you find the range of a particle?
- How would you find the Maximum height reached by a particle?
- How would you find the magnitude of a particle’s velocity after 3 seconds?
- How would you find the direction of a particle after 3 seconds?
- If an object is fired from a cliff which is 200 metres above sea-level, how will this affect our approach to answering the question?
- If a bullet is fired horizontally what does it mean?
- When would we ever take ‘g’ to be positive?
- If two projectiles collide in mid-air then.. .?
Higher Level Projectiles: Projectile is fired up a hill
- Write out expressions for u, a, v, and s in both the i and j directions for a particle fired up a hill. The particle is projected at an initial velocity of 100 m/s at an angle of α to the hill and the hill itself is at an angle of 30^0 to the horizontal.
- If the projectile travels down the plane then.. .?
- If a Projectile lands perpendicular to the slope then.. .?
- If the projectile lands horizontally then.. .?
- How would you find the angle of projection which will result in maximum range?
i-direction j-direction u = u =
a = a =
v = v =
s = s =
Initial velocity is given in terms of i and j components 2010 OL A particle is projected with initial velocity 72 i + 30 j m s-1^ from the top of a straight vertical cliff of height 35 m. It strikes the horizontal ground at P. Find (i) the time taken to reach the maximum height (ii) the maximum height of the particle above ground level (iii)the time of flight (iv) | OP |, the distance from O to P (v) the speed of the particle as it strikes the ground.
2007 OL
A projectile is fired with initial velocity14 i + 10 j m/s from the top of a vertical cliff of height 40 m. (i) Calculate the time taken to reach the maximum height. (ii) Calculate the maximum height of the projectile above ground level. (iii)Calculate the time it takes the projectile to travel from the maximum height to the ground. (iv) Find the range. (v) Find the speed of the projectile as it strikes the ground.
2006 OL
A particle is projected from a point on a level horizontal plane with initial velocity 10 i + 35 j m/s, where i and j are unit perpendicular vectors in the horizontal and vertical directions respectively. Find (i) the time it takes to reach the maximum height (ii) the maximum height (iii)the two times when the particle is at a height of 50 m (iv) the speed with which the particle strikes the plane.
2009 (a) OL A particle is projected with initial velocity 40 i + 50 j m/s from point p on a horizontal plane. a and b are two points on the trajectory (path) of the particle. The particle reaches point a after 2 seconds of motion. The displacement of point b from p is 360 i + kj metres. Find (i) the velocity of the particle at a in terms of i and j (ii) the speed and direction of the particle at a (iii)the value of k.
Initial velocity needs to be resolved into i and j components
Resolving a vector into two perpendicular Components
First we need to remember that for a right-angled triangle: Sin θ = Opposite/Hypothenuse, therefore Opposite = Hypothenuse x Cos θ {Opp = H Sin θ} Cos θ = Adjacent/Hypothenuse, therefore Adjacent = Hypothenuse x Cos θ {Adj = H Cos θ}
Example Consider a velocity vector representing a velocity of 50 ms-1, travelling at an angle of 60^0 to the horizontal: The Opposite is equal to H Sin θ, which in this case = 50 Cos 60^0 = 43 ms-1. The Adjacent is equal to H Cos θ, which in this case = 50 Sin 60^0 = 25 ms-1.
2008 OL
A particle is projected from a point on horizontal ground with an initial speed of 25 m/s at an angle β^0 to the horizontal where tan β = 4/3. (i) Find the initial velocity of the particle in terms of i and j. (ii) Calculate the time taken to reach the maximum height. (iii)Calculate the maximum height of the particle above ground level. (iv) Find the range. (v) Find the speed and direction of the particle after 3 seconds of motion.
2005 (a) OL A particle is projected from a point o on level horizontal ground with an initial speed of 50 √3 m/s at an angle β to the horizontal. It strikes the level ground at p after 15 seconds. (i) Find the angle β. (ii) Find op , the distance from o to p. Give your answer to the nearest metre.
2004 (b) OL A golf ball is struck from a point r on the horizontal ground with a speed of 20 m/s at an angle θ to the horizontal ground. After 2 seconds, the ball strikes the ground at a point which is a horizontal distance of 40 m from r. (i) Find the initial velocity of the ball, in terms of i and j and θ. (ii) Find the angle θ.
2003 OL
A particle is projected from a point p on level horizontal ground with an initial speed of 50 m/s at an angle β to the horizontal, where tan β = 3/4. (i) Find the initial velocity of the particle in terms of i and j.
After 4 seconds in flight, the particle hits a target which is above the ground. (ii) Show that the distance from the point p to the target is 40√17 m. (iii)How far below the highest point reached by the particle is the target? (iv) Find, correct to the nearest m/s, the speed with which the particle hits the target.
Answers to Ordinary Level Exam Questions 2010 (i) t = 3 s (ii) distance = 80 m (iii)t = 7 s (iv) OP= 504 m (v) v = 82.4 m s-
2009 (a) (i) V = 40 i + 30 j (ii) Speed = 50 m s-1, θ = 36.87^0 (iii)k = 45
2009 (b) x = 17.
2008 (i) v = 15 i + 20 j (ii) t = 2 s (iii)s = 20 m (iv) range = 60 m (v) speed = 18.0 m s-1, θ = 33.69^0
2007 (i) t = 1 s (ii) Max height = 45 m (iii)t = 3 s (iv) Range = 56 m (v) Speed = 33.11 m s-
2006 (i) t = 3.5 s (ii) maximum height = 61.25 m (iii)t = 2 s and t = 5 s (iv) Speed = 36.4 m s-
2005 (a) (i) = 60^0 (ii) op= 650 m
2005 (b) u = 129.9 m s-
2004 (a) (i) u = 4 m s- (ii) h = 1 m
2004 (b) (i) Initial velocity = 20 cosθ i + 20 sinθ j (ii) = 45^0
(i) Initial velocity = 40 i + 30 j (ii) (iii)5 m (iv) Speed = 41 m s-
2002 (i) t = 4 s (ii) x = 20 m s- (iii) (iv) Range = 60 m
2001 (i) t = 3 s (ii) Distance = 60 m (iii)y = 22.5 m
2000 (a) S = 45 m T = 6 s pq= 240 m
A ball is kicked from level ground. The first bounce occurs at the point r , 45 m from the kicking point O and the greatest height reached was 22.5 m. If the horizontal and vertical components of the initial velocity are taken as p and q as in the diagram, calculate (i) the value of p and the value of q. (ii) the farthest distance from r that a person running at 7 m/s can be, so that starting when the ball was kicked, the person can be at r just as the ball lands.
1996 (a)
A particle is projected from the ground with a velocity of 50.96 m/s at an angle tan-^ 5 12 to the horizontal. On its upward path it just passes over a wall 14.7m high. During its flight it also passes over a second wall 18.375 m high. Show that the second wall must be not less than 23.52 m and not more than 70.56 m from the first wall.
A particle is projected with velocity 6√ g m/s, at an angle α to the horizontal, from a point 18 m in front of a vertical wall 5. 5 m high. (i) Calculate the two possible values of α which will enable the particle to just clear the wall. (ii) Show that the value of α is tan-12 for maximum clearance height.
2010 (a) In a room of height 6 m, a ball is projected from a point P. P is 1.1 m above the floor. The velocity of projection is 9.8 2 m s-1^ at an angle of 45^0 to the horizontal. The ball strikes the ceiling at Q without first striking a wall. Find the length of the straight line PQ.
2008 (a) A ball is projected from a point on the ground at a distance of a from the foot of a vertical wall of height b , the velocity of projection being u at an angle 45° to the horizontal.
If the ball just clears the wall prove that the greatest height reached is
2004 (a) A particle is projected from a point on the horizontal floor of a tunnel with maximum height of 8 m. The particle is projected with an initial speed of 20 m/s inclined at an angle α to the horizontal floor. Find, to the nearest metre, the greatest range which can be attained in the tunnel.
A bullet is fired from a gun fixed at a point o with speed v m/s at an angle of θ to the horizontal. At the instant of firing, a moving target is 10 m vertically above o and travelling with a constant speed
42 2 m/s at an angle of 45^0 to the horizontal. The bullet and target move in the same plane. (i) If v = 70 m/s and tan θ = 4 / 3 , find at what time after firing does the bullet strike the target and calculate the horizontal distance of the bullet from o. (ii) Show that the minimum value of θ to ensure that the bullet strikes the target is given by tan θ = 4 / 3
Particle fired from a horizontal surface - general questions 2012 (a) A particle is projected with a speed of 98 m s−1^ at an angle α to the horizontal. The range of the particle is 940·8 m. Find (i) the two values of α (ii) the difference between the two times of flight.
2007 (a) A particle is projected with a speed of 7√5 m/s at an angle α to the horizontal. Find the two values of α that will give a range of 12.5 m.
2009 (a) A straight vertical cliff is 200 m high. A particle is projected from the top of the cliff. The speed of projection is14 √10 m/s at an angle α to the horizontal. The particle strikes the level ground at a distance of 200 m from the foot of the cliff. (i) Find, in terms of α, the time taken for the particle to hit the ground. (ii) Show that the two possible directions of projection are at right angles to each other.
2003 (a) A particle is projected from a point on level horizontal ground at an angle θ to the horizontal ground. Find θ, if the horizontal range of the particle is five times the maximum height reached by the particle.
1993 (a) A particle is projected on a horizontal plane with initial velocity u at an angle β to the horizontal. If the range of the projectile is three times the greatest height, prove that tan β = 4 / 3.
1986 A particle is projected with speed u at an angle α to the horizontal. (i) If the maximum height reached is the same as the total horizontal range, show that tan α = 4.
(ii) The particle moves at right angles to its original direction of motion after a time t 1 and then strikes the horizontal plane after 8 s, both times measured from the instant of projection.
Show u = g 17
(iii)Calculate t 1.
2002 (a) A particle is projected from a point on the horizontal ground with a speed of 39.2 m/s inclined at an angle α to the horizontal ground. The particle is at a height of 14.7 m above the horizontal ground at times t 1 and t 2 seconds, respectively.
(i) Show that t 2 – t 1 =
(ii) Find the value of α for which t 2 - t 1 =.
1988 (a)
A particle which is projected with speed u has a horizontal range 49
3 u^2 .
Calculate the two possible angles of projection.
64sin^2 α − 12
A particle is projected from a point O on a plane inclined at 60^0 to the horizontal with a velocity
u = 7 3 i
+ 4.9 j
metres/second where (^) i
is a unit vector through O pointing upward along the line of
greatest slope in the plane and j
is a unit vector perpendicular to the plane.
Show that after time t seconds the position vector, r
of the particle relative to O is given by
r
[(20 3 t – 7 3 t^2 ) i
+ (14 t – 7 t^2 ) j
] metres.
Prove that the range on the inclined plane is 5
(^21 3) metres, and find the velocity of the particle when it strikes
the plane.
(a) A particle is projected under gravity with an initial velocity vo at an angle θ to the horizontal. Find its position and the direction of motion after time T in terms of vo , θ, g and T.
(b) A particle is projected from the top of a cliff which is 425 ft. above sea level and the angle of projection is 450 to the horizontal. If the greatest height reached above the point of projection is 200ft, find the speed of the projection and the time taken to reach this greatest height. Find when and where the particle strikes the sea. (Take g to be 32 ft/sec^2 .)
Maximum Range
1981 (a) Establish an expression, in terms of initial speed u and angle of inclination to the horizontal α, for the range of a projectile on a horizontal plane through the point of projection. Deduce that the maximum range for a given u is u^2 / g.
2000 (a) A particle is projected with a velocity of u m/s at an angle β to the horizontal ground.
Show that the particle hits the ground at a distance β from the point of projection.
Find the angle of projection which gives maximum range.
A particle is projected with speed u at an angle α to the horizontal. The range of the particle on the horizontal plane through the point of projection is R. (i) Show that R is a maximum when α = 45^0.
(ii) If R = g
u 2
2 , find the two possible values of α.
(iii)If the ratio of the greatest height to the range is 2 : 5, find α.
u^2 g
sin 2
Equations of motion for a particle on an inclined plane
x-direction y-direction
u
a
v
s
You need to be familiar with the following trigonometric functions
