DAILY PRACTICE PROBLEMS

PHYSICS

P05

Projectile Motion Topics : Projectile Motion

SECTION - I (Straight Objective Question) This section contains seven (7) multiple choice questions. Each question has four (4) options (a), (b), (c), (d) out of which ONLY ONE is correct. 1. A bullet is fired upwards from the top of a tower with a velocity of 50 ms–1 making an angle of 30º with the horizontal. The height of the tower is 70 m. The time taken by the bullet to strike the ground is — (a) 7 s (b) 5 s (c) 9 s (d) 2 s Sol. (a) Vertical component = u sin  u  50sin 30º  25 ms 1 h  70m  u  25ms 1

 70   25t  

1 2 gt  70  25t  5t 2 2

 t 2  5t  14  0

 t  7t  2t  14  0

  t  7  t  2   0 t = – 2 sec or, +7 sec  t  7 sec 2. A point is at a distance of 15 m from a foot of vertical wall. A ball is thrown at an angle of 45º towards the wall and just clears its top. Then it falls on a ground at a distance of 5m from the wall on other side. The height of the wall is — (a) 5.75 m (b) 4.75 m (c) 3.75 m (d) 2.75 m

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PHYSICS

Sol. (c) Horizontal range = 15 + 5 = 20 m

R

u 2 sin 2  20 g

u2  20 g

  45º

 u 2  20  10

u  10 2 ms 1 Co-ordinates of print p = (15, h) equation of trajectory

y  x tan  

1 gx 2 2 u 2 cos 2 

1 g 15  15 1 1 15  15 45 y  15 tan 45º  . 2  15     15  2 2 u cos 45º 2 20 1/ 2 4

15  3.75m 4  h = 3.75 m 3. An object is projected with a velocity of 20 m/s. making an angle of 45º with horizontal. The equation of trajectory is y = Ax – Bx2, where A and B are constants. The ratio A : B is (a) 1 : 5 (b) 5 : 1 (c) 1 : 50 (d) 40 : 1 Sol. (d) Equation of trajectory is y

y  x tan  

1 x2 g 2 2 v0 cos 2 

 A  tan   1    45º

B

1 1 g 2 2 v0 cos 2 

v0 = 20 m/s 1 1 1 A  10.    40 2 1 2 40 B  20  . 2 A : B = 40 : 1 4. For a given velocity, a projectile has the same range R for two angles of projection. If t1 and t2 are the times of flight in the two cases, then — B 

(a) R 2  t1t2

(b) R  t1t2

1 (c) R  t t

12

1 2 (d) R  t t

12

Sol. (b) For a given velocity of projection there are two angles of projection for which range is same. Sum of these angles is 90º. Let angles be  and  .     90º Then, t1 

2

2 v0 sin  2v sin  and t2  0 g g

Projectile Motion

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PHYSICS

 t1t2 

4 v02 sin  sin  90    g



2v02 sin 2  2R g

 R  t1t2 5. A particle A is projected from the ground with an initial velocity of 10 m/s at an angle of 30º with the vertical. From what height should another particle B be projected horizontally with velocity 5 m/s so that both the particles collide in ground at point Q, when projected simultaneously ? P

h 30º

5 m/s

10 m/s Q

O

(a) 10 m (b) 15 m Sol. (b) Range of projectile projected from ground is

R

(c) 25 m

(d) 30 m

v02 sin 2  (10)2  sin 60º 3   10 5 3 m g 10 2

Range of projectile when projected from height h is R  v

2h 2h 5  5h g 10

 5 3  5h or, h  15m 6. A projectile has velocities 3 m/s and 4 m/s at two points of its flight in the uniform gravitational field of the earth. If these two velocities are perpendicular to each other, then its velocity at the higher point is (a) 2 ms–1 (b) 2.4 ms–1 (c) 3.5 ms–1 (d) 3.8 ms–1 Sol. (b) If direction of velocity makes an angle  at one point, then its direction at other point makes an angle 90   . Horizontal component of velocity remains constant. Hence 3 3 cos   4 cos  90     4sin  or, tan   4 Velocity at highest point is only in horizontal direction.

4  2.4 ms 1 5 7. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m/s can go without hitting the ceiling of the hall ? (a) 150.7 m (b) 140.7 m (c) 145.7 m (d) 160.7 m Sol. (a) Given H = 25 m, u = 40 m/s If the ball is thrown at an angle  with the horizontal, then maximum height of flight is given by Hence velocity  3 cos   3 

H

u 2 sin 2  2g

or, 25 

402 sin 2  2  9.8

Which on solving gives sin   0.554 and cos   0.833 The maximum horizontal distance is given by Projectile Motion

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DPP R

PHYSICS u 2 sin 2 2u 2 sin  cos  2  402  0.554  0.833    150.7 m g g 9.8

SECTION-II (One or More than correct objective Question) This section contains three (3) multiple choice questions. Each question has four (4) options (a), (b), (c), (d) out of which ONE OR MORE is/are correct. 8. The projection speed of a projection is 10 ms–1. Neglecting air effects and taking g = 10 ms–2, its horizontal range is found to be 5m. Calculate the angle(s) of projection. (a) 15º (b) 60º (c) 75º (d) 45º Sol. (a, c) Given, R 

u 2 sin 2 g

Here, u = 10 ms–1, R = 5m, g = 10 ms–2.

10 ms 1  sin 2 5 10 ms 2  2

1 2  2  30º   15º Since the horizontal range is less than maximum, there are two angles of projection for the same horizontal range of 5m. One of them we have found as 15º. As the sum of the two angle of projection for the same horizontal range is 90º, the other angle will be 90º – 15º = 75º. Thus, there are two angles of projection, 15º and 75º for the same horizontal range of 5m. 9. A projectile has same range R for two angles of projection. If T1 and T2 are the times of flight in the two cases, then — sin 2 

(a) T1 T2  R

T1 (b) T  tan  2

(c) T1 T2  R 2

T1 (d) T  cot  2

Sol. (a, b) Sum of the two angles of projection is 90º. 2V sin  2V cos  T1  1 and T2  1 g g

T1T2 

4V12 sin  cos   2R g

 T1T2  R

and

T1  tan  T2

10. Shots are fired simultaneously from the top and bottom of a vertical cliff at angles  , and  and they strike an object simultaneously at the same point. Show that if the horizontal distance of the object from the cliff is a, the height of the cliff is —

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Projectile Motion

DPP (a)

PHYSICS a tan  tan 

(b) a  cot   cot  

(c)

a  cot   cot   cot  cot 

(d) a  tan   tan  

Sol. (c, d) Co-ordinates of P w.r.t. O and O´ as origins are (a, n) and (a, –m). The height of cliff = h = m + n y  x tan  

gx 2 2u 2 cos 2 

  m  a tan  

ga 2 2u cos  2

2

 n  a tan  

ga 2 2v 2 cos 2 

 n  m  a  tan   tan   because u cos   v cos  because both the shots reach P simultaneously. This is possible only if the horizontal components of the velocities are equal.  1 1  a  cot   cot   h  a  tan   tan    a    cot .cot   cot  cot  

SECTION - III (Assertion & Reasoning Type Question)

(a) (b) (c) (d) 11.

This section contains three (3) assertion & reasoning type questions. Each question has four (4) options (a), (b), (c), (d) out of which ONLY ONE is correct. Statement based Questions : For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows ; Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I. Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I Statement I is true; Statement II is false. Statement I is false; statement II is true. STATEMENT-I 2 The maximum possible height attained by the projected body is u , where u is the velocity of 2g

projection. STATEMENT-II To attain the maximum height, body is thrown vertically upwards. Sol. (a) For maximum height   90º , or body must be projected straight upwards. Then 0  u 2  2 gh .

u2 2g 12. STATEMENT-I When the range of projectile is maximum, the time of flight is the largest. STATEMENT-II Range is maximum when angle of projection is 45º. Sol. (d) h 

T

2u sin  , it will maximum, when   0º. g

Projectile Motion

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PHYSICS

Rmax 

u2 , for   45º g

13. STATEMENT-I Two projectiles having same range must have the same time of flight. STATEMENT-II Horizontal component of velocity is constant in projectile motion under gravity. Sol. (d) Statement-I is false because angles of projection  and (90º –  ) give same range but time of flight will be different. Statement-II is true because in horizontal direction acceleration is zero.

SECTION - IV (Passage Based Question) This section contains one (1) paragraph based upon the paragraph three(3) multiple choice questions have to be answered. Each question has four(4) options (a), (b), (c), (d) out of which ONLY ONE is correct.

Passage A particle is projected from the surface of the earth with a speed of 20 m·s–1 at an angle 30º with the horizontal. (g = 10) 14. The time of flight of that particle is (a) 3 s (b) 4 s (c) 2 s (d) 1 s Sol. (c)

T

2u sin  2  20  sin 30º   2s g 10

15. The range of that particle is (a) 10 m

(b) 12 2 m

(c) 20 3 m

(d) 30 m

Sol. (c) Range R 

u 2 . sin 2 20  20  sin 60º 3   40   20 3 g 10 2

16. The maximum height the particle can reach is (a) 3 m

u 2 sin 2  Sol. (c) H   2g

(b) 7 m

(c) 5 m

(d) 12 m

1 4  400  5 m 2  10 80

20  20 

SECTION - V (Matrix Match Type) This section contains one (1) question. It contains statements given in two columns, which have to be matched. Statements in column I are labelled as A, B, C and D whereas statements in column II are labelled as p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A  p, r ; B  q, s; C  r and D  q, then the correctly bubbled matrix will look like the following: 6

A B C D

p q

r

s

p p p p

r r r r

s s s s

q q q q

Projectile Motion

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PHYSICS

17. The equation of trajectory of a particle projected from the surface of the planet is given by the equation y = x – x2 . Then match the following columns : (suppose, g = 2 m/s2) Column I Column II (magnitude only)

1 4 (q) 1/2 (r) 1/3 (s) 1

(a) angle of projection, tan 

(p)

(b) T/3 (c) maximum height attained, H (d) R/2 Sol. (a)  s ; (b)  r ; (c)  p ; (d)  q Given, y = x – x2 . . . (i)

y  x tan  

g x2

. . . (ii) 2 u 2 cos  Comparing equation (i) and (ii) we get, tan   1  tan 45º   45º g 1 Again, 2 2u cos 2 45º u2  2  u  2 (a) angle of projection = tan  = 1

T 2u sin  (b)   3 3g

2 2  6

1 2 1 3

u 2 sin 2  (c) Maximum height H =  2g

1 21 4 4

2

R u 2 sin 2 2  sin90º 1    (d) 2 2g 4 2

SECTION - VI (Integer Type) This section contains three(3) questions. The answer to each of the question is a single digit integer, ranging from 0 to 9. 18. A soccer player kicks a ball at an angle 30º from the horizontal with the initial speed of 20 m/s. The ball moves in a vertical plane. Calculate the maximum vertical hight attain by the ball. (g = 10 m/s2) Sol. (Ans. 5) y At t = 0, the player strikes the ball. Therefore, at t = 0. vx(0) = 20 cos30º vy(0) = 20 sin 30º Time to reach maximum height is given by ymax 30º 0 = 20 sin 30º – gt, [ at the maximum height vy = 0] x or, t  20 

sin 30º  1 second g

1 2  ymax  20 sin 30º 1   10  1  5 m 2 19. A projectile is thrown with an initial velocity of x iˆ  y ˆj . The range of the projectile is twice the maximum height of the projectile. Calculate

y . x

Sol. (Ans. 2) Projectile Motion

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PHYSICS

v 2 sin 2 v 2 sin 2  2 g 2g or, 2v 2 sin  cos   v 2 sin 2  or, 2  v sin   v cos     v sin    v sin   But v sin   y and v cos   x

 2 yx  y 2 or, 2 x  y or,

(given)

y 2 x

20. For the angles of projection of a projectile at angles  45º   and  45º   the horizontal range are R1 and R2 respectively. Calculate Sol. (Ans. 1) The horizontal range, R 

R1 . R2

u 2 sin 2 g

For the projectile fired at (45º ), the horizontal range,

R1 

2 u 2 sin 2(45º ) u sin  90º 2   g g

or, R1 

u 2 cos 2 g

. . . (i)

For the projectile fired at (45º ), the horizontal range,

R2 

2 u 2 sin 2(45º ) u sin  90º 2   g g

or, R2 

u 2 cos 2 g

. . . (ii)

Combining equation (i) and (ii) we get Hence,

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R1 1 R2

Projectile Motion