7953m d h మ మ ఏ ? ? మ ௦? మ ? ?

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= 79.53m d.) H = ௌ௜௡ ଶ௚ = ଷ଴ ௦௜௡ଷ଴ ଶ ௫ ଽ.଼ଵ H =11.48m 2.From the edge of the cliff 60m high, a stone is thrown into air with speed of 10m/s at an angle of 30 o to the horizontal as shown below.
Prepared By: ChilesheMwenyaMporokoso Page 17 Find; a.)The time of flight b.)How far from the foot of the cliff does it strike the sea c.)the velocity with which it strikes the sea. Solution Taking the upward position above the throwing point as positive and g(acceleration due to gravity) when stone is going up as negative a.)Using H = Ut + gt 2 We have 60 =Usin30 x9.81xt 2 60 = 10x0.5t 4.9t 2 4.9t 2 5t – 60 = 0 This is a quadratic equation Using the formula x = ି௕േ√௕ ିସ௔௖ ଶ௔ t = ହ േඥହ ି ସሺସ.ଽሻሺିሺ଺଴ሻ ଶ ௫ ଽ.଼ t = 4.05sec OR t = 3.03 this is invalid Thus t=4.05sec b.) R =Utcos θ R = 10x 4.05 x cos30 0 R = 35.07m c.)V y = usin30 – gt from the equation v = u +at V y = 10sin30 9.8 x 4.05 V y = 34.69m/s V x = Ucos30
Prepared By: ChilesheMwenyaMporokoso Page 18 10x cos30 =8.66m/s V R = √8.66 ൅ 34.69 V R = 35.75m/s This is the velocity with which the stone hits the ground. 3.An arrow is short with a velocity of 30m/s at an angle of 37 o above the horizontal. The arrow is initially 2.0m above the ground and 15.0m from the wall as shown below. a.)At what height above the ground does it hit the wall? b.)Is it going up just before it hits the wall or is it already on its way down? Solution When the arrow is going up the quantities velocity, acceleration due to gravity and height are given the following signs Velocity = +Ve Gravitational acceleration = Ve Height = +ve When the arrow is going down the quantities velocity, acceleration due to gravity and height are given the following signs
Prepared By: ChilesheMwenyaMporokoso Page 19 Velocity = Ve Gravitational acceleration = +Ve Height = ve Using the Trajectory equation Y = xtan ߠ ଶ௨ ஼௢௦ ቁ ݔ X= 15m, u= 30m/s, θ = 37 o Y = xtan ߠ ଶ௨ ஼௢௦ ቁ ݔ = 15 tan37 0 ଽ.଼ ଶ ௑ଷ଴ ௑ሺ௖௢௦ଷ଻ሻ (15) 2 = 9.38m b.)Since Y = +ve, the arrow is going up. Self Evaluation Exercise 1.A particle is propelled with a speed of 20m/s and reaches its greatest height above the point of projection 1 second later. Find; (i)the angle of projection (ii)the greatest height 2.A golfer drives a ball horizontally at 60m/s from the top of a cliff 45m above the sea. a.)How long does it take the ball to hit the water? b.)How far from the cliff foot does the ball hit the water? c.)Calculate the velocity of the ball just before it hits the water
Prepared By: ChilesheMwenyaMporokoso Page 20 UNIT 1.4: ANGULAR (ROTATIONAL) MOTION To describe the motion of an object along a line, we need a coordinate along the line, which we often take to be the x or y coordinates. To describe the motion of an object on a circular path or rotation of a wheel on an axis, we need a coordinate system to measure angles. They are three common ways in which angles are measured. These are: (i)Degrees (ii)Radians (iii)Revolutions (iv)Steradians Consider a wheel rotating about an axis O In moving from position A to B, the wheel has turned through the angle θ .

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