Engineering Mechanics

(^408) „„„„„ A Textbook of Engineering Mechanics = 1024++ 576 1536 cos (90°+°5 ) = 1600 – 1536 sin 5° ...[ cos (90Q °+θ =−) sin ...

Chapter 19 : Relative Velocity „„„„„ 409 (b) Bearing of the submarine Fig. 19.9. Again draw East, West, North and South lines m ...

(^410) „„„„„ A Textbook of Engineering Mechanics (a) Actual velocity diagram (b) Relative velocity diagram Fig. 19.10. First of ...

Chapter 19 : Relative Velocity „„„„„ 411 From the geometry of the figure, we find that in triangle XPQ, XQ = 18.1 m, QP = 200 m ...

(^412) „„„„„ A Textbook of Engineering Mechanics Now in order to find the least distance between the two bodies, first of all, w ...

Chapter 19 : Relative Velocity „„„„„ 413 Now cut off XN equal to 15 km to the scale on the North line to represent the actual d ...

(^414) „„„„„ A Textbook of Engineering Mechanics Consider two bodies A and B moving with velocities vA and vB respectively. Let ...

Chapter 19 : Relative Velocity „„„„„ 415 Solution. Given : Velocity of ship A = 12 km.p.h. (North) ; Velocity of ship B = 20 km. ...

(^416) „„„„„ A Textbook of Engineering Mechanics EXERCISE 19.3 Two ships A and B, at a given instant, are 5 km away from each o ...

Chapter 20 : Projectiles „„„„„ 417 417 Projectiles 20 CHAPTERCHAPTERCHAPTERCHAPTERCHAPTER 20.1.INTRODUCTION In the previous chap ...

(^418) „„„„„ A Textbook of Engineering Mechanics direction. The combined effect of both the components is to move the particle a ...

Chapter 20 : Projectiles „„„„„ 419 We know that height of aircraft (s) 1000 112220 9.8 4.9 22 =+ut gt =+×t =t or^2 1000 204.1 4. ...

(^420) „„„„„ A Textbook of Engineering Mechanics Inclination and magnitude of the velocity of motor cycle just after clearing th ...

Chapter 20 : Projectiles „„„„„ 421 We know that the vertical distance OA (s), 1000 –^11222 295 – 9.8 295 – 4.9 y 22 ==×=ut gt t ...

(^422) „„„„„ A Textbook of Engineering Mechanics 20.5.EQUATION OF THE PATH OF A PROJECTILE Fig. 20.5. Path of a projectile. Cons ...

Chapter 20 : Projectiles „„„„„ 423 We know that when the particle is at A, y is zero. Substituting this value of y in the above ...

(^424) „„„„„ A Textbook of Engineering Mechanics 20.8.MAXIMUM HEIGHT OF A PROJECTILE ON A HORIZONTAL PLANE We have already discu ...

Chapter 20 : Projectiles „„„„„ 425 Solution. Given : Horizontal range (R) = 3 H (where H is the greatest height). ...(i) Let α = ...

(^426) „„„„„ A Textbook of Engineering Mechanics Fig. 20.6. (c) Velocity of the second particle for equal time of flight We know ...

Chapter 20 : Projectiles „„„„„ 427 (^22) 2(1cot ) 2 1^1 tan 2 4 ugh=+α= + θgh⎛⎞⎜⎟ ⎝⎠ ... Substituting cot^221 tan 4 ⎛⎞ ⎜⎟α= θ ⎝⎠ ...