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40 PHYSICS range,R.Itisthedistancetravelledduringthetime y (t) = o + oy t +(1/2) ayt2 of flight T . Therefore, the range R is Here, x = y = 0, v = 0, a = –g = –9.8 m s , -2 f o o -1 y R = (vocos θo) (f ) vox 15 m s . The stone hits the ground when y(t) = – 490 m. =(vocos θo) (2 o sin θo/g 2 – 490 m = –(1/2)(9.8) t . v2 sin 2θ This gives t =10 s. Or, R = 0 0 (3.42a) The velocity components are v = v and g x ox vy= voy g t Equation (3.42a) shows that for a given so that when the stone hits the ground : –1 projection velocity vo, R is maximum when sin vox 15 m s 2θ 0is maximum, i.e., when θ0= 45 . voy 0 – 9.8 × 10 = – 98 m s –1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is v2 2 2 2 2 −1 R m = 0 (3.42b) vx+ v y 15 + 98 = 99 m s ⊳ g ⊳ Example 3.6 Galileo, in his book Two new ⊳ Example 3.8 A cricket ball is thrown at a speed of 28 m s in a direction 30° above sciences, stated that “for elevations which exceedorfallshortof45°byequalamounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the Answer For a projectile launched with velocity distance from the thrower to the point where the ball returns to the same level. voat an angle θo , the range is given by 2 v0sin2 θ0 Answer (a) The maximum height is given by R = g 2 (v0 sinθo) (28sin30 °)2 Now, for angles, (45° + α) and ( 45° – α), 2θ is hm = = m o 2g 2 (.8 ) (90° + 2 α) and ( 90° – 2 α) , respectively. The values of sin (90° + 2 α ) and sin (90° – 2α ) are 14 ×14 the same, equal to that of cos 2 α. Therefore, = = 10.0 m 2× 9.8 ranges are equal for elevations which exceed or (b) The time taken to return to the same level is fall short of 45° by equal amounts α . ⊳ Tf= (2 o sinθ o)/g = (2× 28 × sin 30°)/9.8 ⊳ = 28/9.8 s = 2.9 s Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed of 15 m s . Neglecting air resistance, ( oin2 θo ) 28 × 28 × sin60 o find the time taken by the stone to reach R = = = 69 m ⊳ g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s ).2 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- axis at the edge of the cliff and t = 0 s at the When an object follows a circular path at a constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the refers to the speed, which is uniform (constant) throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = o + oxt object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2026-27