Download presentation

Published byLuc Mangin Modified over 3 years ago

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 kmph 16

2
**Let 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.

3
**Therefore, for the motion AB downwards**

for the motion from C to B,

4
**A 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 = ?

5
**A 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,

6
**Let 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,

7
**Let 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.

8
**The 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.

11
**A 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

12
**Distance travelled in 4s is**

(t = 4, u = 0, a = 20 m/s2) Velocity at the end of 4s is v = u + at = 20 x 4 = 80 m/s.

13
**At 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 = 0 From the equation v2 = u2 + 2gs, we get, Let t be time taken to reach the maximum height. Then,

14
**Velocity 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 + at 20 = 0 + a x 10 or a = 2 Let 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

15
Determine 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.

16
**A 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 = 100 For the body dropped For the other body

17
A 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 x Therefore, 8 = 5x

18
A 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 = 4s The time taken to reach maximum height = t = 2s g = 9.8 m/s2 v = 0 v = u + gt v = u + gt 0 = u x 2 u = 19.6 m/s The greatest height reached is

19
A 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.

20
Given , 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.

21
**A 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 = -2gS v = -2 x 9.8 x 10.2 v2 = = 350 v = 18.7 ms-1

22
**A 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

23
**A 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

24
**For the bullet moving vertically upwards**

From equation (1) and (2), we get,

25
A 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) Diagram Time taken by the first stone to reach the maximum height is Distance travelled by first stone in 2s.

26
**Let 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 = 0 For the stone projected vertically upwards From equation (1) + equation (2), we get, 20 = 20t t = 1s Substituting in equation (1) 20 - x = 5 x1 x = 15 m.

27
A 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/s2 Let 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

28
**At 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 = 0 Let h' be the height covered by the rocket when it moves under gravity. Therefore, total height reached by the rocket is h + h' = = = km.

29
**A 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 this displacement = - u Acceleration a = g Time taken = t second. g = 10 ms-2 Let h be the height of the tower.

30
For the stone A, For the stone B,

31
A 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/s2 Net displacement = h = height of the tower.

Similar presentations

OK

Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 4) Linear Motion (Constant Acceleration) Mechanics M1.

Copyright © 2015 Chris J Jewell 1 Mechanics M1 (Slide Set 4) Linear Motion (Constant Acceleration) Mechanics M1.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on social issues in today's society Ppt on revolt of 1857 lord Ppt on live line maintenance techniques Ppt on manual metal arc welding Ppt on arbitration and conciliation act 1996 Ppt on next generation 2-stroke engines Ppt on horizontal axis wind turbine Ppt on hindu religion pictures Ppt on renewable resources Ppt on data collection methods for qualitative research