Presentation on theme: "(i) its uniform retardation, (ii) time taken to come to rest,"— Presentation transcript:
1(i) its uniform retardation, (ii) time taken to come to rest, A car moving with a speed of 18 kmph comes to rest, when it moves through a distance of 100m. calculate(i) its uniform retardation, (ii) time taken to come to rest, (iii) its average speed.Given u = 18 kmph16
2Let the two stones meet at a height x above the ground level. A stone is dropped from the top of a tower 96 m tall. At the same instant another stone is thrown vertically upwards from the foot of the tower with a velocity of 24 ms-1. Where and when do the two stones meet?Let the two stones meet at a height x above the ground level.
3Therefore, for the motion AB downwards for the motion from C to B,
4A body having an initial velocity of 0 A body having an initial velocity of 0.9 ms-1 and uniform retardation, passes through the starting point in 18 seconds. Calculate its retardation.Given, u = 0.9 m/s2, t = 18s, a = ?
5A stoone travels through a distance of 4. 9 m in 0 A stoone travels through a distance of 4.9 m in 0.1 s, and 4,998 m in the next 0.1 s. Calculate g at the place.s1 = 4.9 m, s2 = m,
6Let s1 be the distance covered in time t, starting from rest. Then, A particle starting from rest and moving with uniform acceleration, covers distances s1, s2 and s3 in three consecutive equal intervals of time. Calculate the ratio s1: s2 : s3.Let s1 be the distance covered in time t, starting from rest. Then,
7Let the time taken for the stone to reach the water be t second. A stone is dropped into a well and the sound of splash is heard after 3.91 s. If the depth of the well is 67.6 m , find the velocity of sound.Let the time taken for the stone to reach the water be t second.
8The initial velocity and acceleration is A body moving with uniform acceleration, covers 11 m during the 4 th second and 15 m during the 6th second. Calculate its initial velocity and acceleration.The initial velocity and acceleration is
9(b) the height from which the body falls. A freely falling body covers half of the total distance during the last second of its travel, find(a) the time of fall and(b) the height from which the body falls.Let h be the height from which the body falls in a time t.Then, (h/2) is the distance travelled in (t -1) second.
11A body starting from rest covers a distance of 70 m in the 4th second A body starting from rest covers a distance of 70 m in the 4th second. Find(i) the distance travelled in 4s,(ii) velocity at the end of 3s.Given, u = 0, s3 = 70m, n = 4
12Distance travelled in 4s is (t = 4, u = 0, a = 20 m/s2)Velocity at the end of 4s isv = u + at = 20 x 4 = 80 m/s.
13At the position of maximum height s, v = 0 A body is thrown up with a velocity of 78.4 ms-1. Find how high it will rise and how much time it will take to return to its point of projection?At the position of maximum height s, v = 0From the equation v2 = u2 + 2gs, we get,Let t be time taken to reach the maximum height.Then,
14Velocity attained at the end of 5 s is A body starting from rest accelerates uniformly to attain a velocity of 20 ms-1 in 10 s. Compare the distance travelled by it in first 5 s and next 5s. What is the distance travelled by it in the last second of its motion?V = u + at20 = 0 + a x 10 or a = 2Let s1 and s2 be the distances covered by it in the first and next 5 s respectively.Velocity attained at the end of 5 s is
15Determine the initial velocity and the acceleration of a particle travelling with uniform acceleration in a straight line if it travels 55 m in the 8th sec and 85 m in the 13th sec. of its motion.
16A body is dropped from the top of a tower of height 100 m A body is dropped from the top of a tower of height 100 m. Simulataneously another body is thrown vertically upwards with a velocity of 25 ms-1. When will the two bodies meet?Let the bodies meet t seconds after the bodies begin their journeys.If s1 and s2 be the distances moved by the two bodies, then s1 + s2 = 100For the body droppedFor the other body
17A bullet moving with a velocity of 80 m/s can just penetrate 5 wooden planks of equal thickness. How many such planks can it penetrate if the initial velocity is 124 m/s?Let the thickness of each plank be xTherefore, 8 = 5x
18A stone thrown vertically up is back in the hand of the thrower at the end of 4 seconds. If 'g' at the place is 9.8 m/s2, Calculate the velocity of the throw and the greatest height reached by the stone.Total time = 4sThe time taken to reach maximum height = t = 2sg = 9.8 m/s2v = 0v = u + gtv = u + gt0 = u x 2u = 19.6 m/sThe greatest height reached is
19A stone is dropped from an aeroplane flying horizontally at 100 ms-1, when it is vertically above a point A on the ground. If the stone reaches a point B on the ground, calculate the distance AB. Height of the aeroplane is 78.4 meters from the ground. (g = 9.8 ms-2)O is the point of projection of the stone and OA = 78.4 m. OB is the actual path along which the stone moves, until it reaches the ground. But the time taken by the stone to reach the earth is equal to that taken by a body to move along the vertical and cover the path OA.
20Given , u = 0, s = 78.4, g = 9.8, t = ?The horizontal velocity of the stone is uniform.s = ut = 100 x 4 = 400m... AB = 400m.
21A body is projected vertically upwards with a velocity of 20 ms-1 A body is projected vertically upwards with a velocity of 20 ms-1. Calculate(i) maximum height reached,(ii) time taken to reach the maximum height,(iii) velocity midway during the upward journey.(i) v2 - u2 = -2gS(ii) v = u - gt(iii) u = 20, S = 10.2, v = ?v2 - u2 = -2gSv = -2 x 9.8 x 10.2v2 = = 350v = 18.7 ms-1
22A stone from rest, describes, 34. 3 m in the last second of its fall A stone from rest, describes, 34.3 m in the last second of its fall. Calculate the height from which the stone fell and the time of fall. (g = 9.8 ms-2)Let n be the time of fall. Then
23A wooden block is dropped from the top of a cliff 200 m high A wooden block is dropped from the top of a cliff 200 m high. Simultaneously a bullet is fired from the foot of the cliff upwards with a velocity of 200 m/s. Where and after what time will they meet?Let x be the height at which the wooden block and bullet meet after 't' seconds.For the block moving vertically downwards
24For the bullet moving vertically upwards From equation (1) and (2), we get,
25A stone is projected vertically upwards from a point on the ground with a velocity of 20 m/s. Two seconds later another stone is projected from the same point with the same velocity. When and where do they meet? (g = 10 m/s2)DiagramTime taken by the first stone to reach the maximum height isDistance travelled by first stone in 2s.
26Let the two stones meet at a distance x from the ground after a time 't' second. From the maximum height, the first stone starts falling down. (Now g = 10 m/s2, s = 20 - x, u = 0For the stone projected vertically upwardsFrom equation (1) + equation (2), we get, 20 = 20t t = 1sSubstituting in equation (1)20 - x = 5 x1 x = 15 m.
27A rocket is fired vertically from the ground with resultant vertical acceleration of 10 m/s2. The fuel is finished in 60 seconds and it continues to move up. What is the maximum height reached?Initial velocity of the rocket, u = , a = 10 m/s2Let h be the height covered by the rocket in t = 60s.Let v be the final velocity at the end of 60 s.... v = u + at = x 60 = 600 m/s
28At the end of 60 s, the rocket moves with an initial velocity of 600 m/s under gravity. ... u = 600 m/s. a = m/s2, v = 0Let h' be the height covered by the rocket when it moves under gravity.Therefore, total height reached by the rocket is h + h'== = km.
29A stone is dropped from the top of a tower A stone is dropped from the top of a tower. Another stone B is thrown downwards from the same point 2s later with an initial speed of 30 ms-1. If both the stones A and B reach the ground together, find the height of the tower.Consider a body thrown upwards from the top of the tower with a velocity u, it reaches the ground in 't' seconds.Net displacement = h = height of the tower.Initial velocity for thisdisplacement = - uAcceleration a = gTime taken = t second.g = 10 ms-2Let h be the height of the tower.
31A stone is projected vertically up from the top of a tower at the velocity of 9.8 m/s. It reaches the foot of the building in 5 seconds. What is the height of the tower and velocity of the stone when it reaches the ground? (g = 9.8 m/s2)u = m/s, t = 5s, g = 9.7 m/s2Net displacement = h = height of the tower.