# It maintains this speed for 30 seconds and it is then

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It maintains this speed for 30 seconds and it is then brought to rest at a uniform rate in 540m. For the motion of this car;l (i)Draw the velocity – time (ii)Determine the uniform acceleration (iii)Determine the uniform retardation (iv)Calculate the total displacement for the whole journey 2. A ball is thrown vertically upwards at a velocity of 80m/s from the top of a 90m high building. Calculate; (i)maximum height reached by the ball above the building (ii)time taken to reach the maximum height (iii)velocity of the ball just before hitting the ground (iv)duration of the whole motion
Prepared By: ChilesheMwenyaMporokoso Page 14 UNIT 1.3: PROJECTILE MOTION Definition:A projectile is an object propelled into air at an angle to the horizontal and moves under the influence of its own weight. Examples of projectiles a.)A golf ball in flight b.)A tennis in flight c.)A bullet in flight Consider a ball that is thrown with an initial velocity u at an angle θ to the horizontal The path taken by a projectile is called the trajectory. The angle θ the projectile makes with the horizontal is called the angle of projection. The distance the projectile covers along the horizontal direction is called the Range. As the projectile moves upwards, it covers both horizontal as well as vertical distances. We now look at the horizontal distance. HORIZONTAL COMPONENTS There is no horizontal component of acceleration in the horizontal direction. The velocity in the horizontal direction is constant,i.e. From V x =U x + at Since a x = 0 and U x = Ucos θ , then V x = Ucos θ …………………………………………………..(i) The distance moved in the x direction is given by X = V x t, where t is the time of flight Thus x = Utcos θ
Prepared By: ChilesheMwenyaMporokoso Page 15 OR R = Utcos θ ………………………………………….(ii) Where R is called the range . VERTICAL COMPONENTS At maximum height, B, V y = 0 From V = U + at when U = Usin θ and a = g We have 0 = Usin θ ‐ gt t m = ௎ௌ௜௡ఏ ………………………………(iii) t m is the time taken to reach maximum height TIME OF FLIGHT This is the time taken to complete the whole journey of flight from point A to point C. Since t m = ௎ௌ௜௡ఏ from A to B But time taken to move from A to B is equal to time taken to move from B to C Therefore total time of flight is 2t m t = ଶ௎ௌ௜௡ఏ ………………………………………(iv) MAXIMUM HEIGHT REACHED From S=H = U x t m + a(t m ) 2 And when U x =sin θ t = ௎ௌ௜௡ఏ We have H =UxUsin θ X ௎ௌ௜௡ఏ g ௎ௌ௜௡ఏ 2 H = ௌ௜௡ ଶ௚ ………………………………………………(v)
Prepared By: ChilesheMwenyaMporokoso Page 16 Using t = ௎ௌ௜௡ఏ in the equation R = Utcos θ We have R = ௌ௜௡ଶఏ …………………………………………..(vi) Examples 1.A particle is propelled from O with speed of 30m/s at an angle of 30 0 to the horizontal. Find: a.) The time of flight b.) Range of flight c.) The maximum height reached Solution a.) t = ଶ௎ௌ௜௡ఏ = ଶ ௫ ଷ଴ ௫௦௜௡ଷ଴ ଽ.଼ଵ t = 3.06s b.) R = ௌ௜௡ଶఏ = ଷ଴ ௌ௜௡ሺଶ ௫ ଷ଴ሻ ଽ.଼ଵ

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