TJC Unit 5 Projectile Motion

i Unit 5: Projectile Motion Learning Outcomes (a) describe and use the concept of weight as the force experience by a mass in a gravitational field. (b) describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction. (c) derive, from the definition of work done by a force, the equation ∆Ep = mg∆h for gravitational potential energy changes in a uniform gravitational field (e.g. near the Earth’s surface). (d) recall and use the equation ∆Ep = mg∆h to solve problems. (e) describe qualitatively, with reference to forces and energy, the motion of bodies falling in a uniform gravitational field with air resistance, including the phenomenon of terminal velocity. Name : ________________________________ Class : _________

5.1 Weight LO(a) In unit 3, we saw that all objects released near the surface of the Earth fall with the same acceleration (the acceleration of free fall) if air resistance can be neglected. The force causing this acceleration is the gravitational attraction of the Earth on the object. The gravitational force which acts on an object is called the weight of the object. We can apply Newton’s second law to the weight. For a body of mass m falling with the acceleration of free fall g, the weight W is given by W mg= The SI unit of force is the newton (N). This is also the unit of weight. Because weight is a force and force is a vector, we ought to be aware of the direction of the weight of an object. It is towards the centre of the Earth. mass weight a measure of the inertia of a body a measure of the gravitational force on a body SI unit: kg SI unit: N same regardless of location varies with gravitational field strength at each location Example 1 A boy throws a ball vertically upwards. It rises to a maximum height, where it is momentarily at rest, and falls back to his hands. Which of the following gives the acceleration of the ball at various stages in its motion? Take vertically upwards as positive. Neglect air resistance. rising at maximum height falling A −9.81 m s-2 0 +9.81 m s-2 B −9.81 m s-2 −9.81 m s-2 −9.81 m s-2 C +9.81 m s-2 +9.81 m s-2 +9.81 m s-2 D +9.81 m s-2 0 −9.81 m s-2 Weight is the force experience by a mass in a gravitational field.

5.2 Projectile Motion LO(b) So far we have been dealing with motion along a straight line; that is, one-dimensional motion. We will now look at the motion of particles moving in paths in two dimensions. The stroboscopic photographs below depicts three balls A, B and C released simultaneously. Ball A Ball B Ball C Ball A is dropped freely while ball B is projected horizontally. Ball C is projected at an angle to the horizontal. Notice that in equal time intervals balls A and B fall the same vertical distance. This is because both balls undergo the same vertical acceleration due to gravity. Balls A and B also reach the ground at the same time! The paths or trajectory of balls B and C are parabolic. Ball A is an example of a one-dimensional linear accelerated motion whereas balls B and C are said to undergo projectile motion, which is a two-dimensional motion under a constant force. A projectile motion can be considered as a combination of two independent motions. 1. Horizontal motion with uniform velocity (since there is no force in horizontal direction). 2. Vertical motion with uniform acceleration (due to force of gravity).

Horizontal and vertical motion of a projectile • Projectile motion involves motion in the x and y directions simultaneously. • For non-linear motion, the direction of velocity at each point of its trajectory is tangent to the trajectory at that point. • At each point , the velocity of the projectile can be resolved into its horizontal and vertical components respectively. • • Components of velocity at any one point; vX and vy y x θ u ux uy ay ux ux ux ux ux x y u uy ux  At time t = 0, when velocity = u, the magnitudes of the components are given by: ux = u cos  uy = u sin  If substituting the components of velocity into equation of motion, you have to apply the sign (- or +) for the initial velocity according to the sign convention and direction of velocity. x: uniform velocity in the horizontal (x direction) y: uniform acceleration (downwards) due to gravity

• If both the components are known, both the magnitude and direction of the vector can be found. Magnitude of vector: 22 xyu u u=+ Direction of vector: tan y x u u = in this way, the angle of projection can be found. 5.2.1 Analysis of projectile motion Key terms used in projectile motion • Trajectory – the path taken by a projectile • Range, R – the distance on the plane between the point of projection and the point of impact • Angle of projection – the angle between the direction of projection and the horizontal plane through the point of projection • Time of flight – time taken from the point of projection to the point of impact Strategies used in solving problems involving projectile motion without air resistance 1. If diagram of path is not given, sketch one, identifying start and end point. 2. Resolve given velocity into its x and y components. 3. Apply the relevant equations of motion separately to the horizontal and vertical motion. (Remember to take sign conventions into consideration!) Horizontal direction, x Vertical direction, y ux = ? uy = ? vx = ux (ax = 0) vy = uy + ayt sx = uxt (ax = 0) sy = uyt + ½ ayt2 vy2 = uy2 + 2 aysy

Projectile launched horizontally Horizontal direction Vertical direction Initial velocity (Starting point is the highest point) ux = u The horizontal component of the object’s velocity remains unchanged throughout its flight. uy = 0 The vertical component of the object’s velocity increases as the object falls towards ground. Sign convention* right positive + downward positive + Acceleration ax = 0 ay = 9.81 m s−2 (taking downwards as positive) Component of velocity at any given time, t. vx = ux = u vy = uy + ayt = gt since uy = 0 Time of flight T Consider the vertical motion; using sy = uyt + ½ ayt 2 and taking the downward direction as positive, h = 0 + ½ gT2 T = g h2 Displacement horizontal range = sx Using sx = uxt + ½ axt2 since ax = 0, sx = uxt = ut Range, R = uT = u g h2 vertical displacement = sy Using sy = uyt + ½ ayt2 since uy = 0, ay= g sy = ½ gt2 By combining the two equations, we have y = 2 2g(x/u) = 22u g x2. This implies y  x2. Thus the trajectory of a projectile is a parabola.

Example 2 A 50 g ball is thrown horizontally from a cliff top with a horizontal speed of 10 m s-1. If the cliff is 20 m high, calculate (a) the magnitude and direction of the velocity of the ball when it strikes the sea, Solution: Horizontal component of velocity: Vertical component of velocity: Final velocity: The ball strikes the sea at __________ m s-1 at angle ___________ below the h