Q 1 :

The angle of projection for a projectile to have same horizontal range and maximum height is                       [2024]

  • tan-1(2)

     

  • tan-1(12)

     

  • tan-1(4)

     

  • tan-1(14)

     

(3)  

        Hmax=Range

       u2sin2θg u2sin2θ2g

        4sinθ cosθ=sin2θ

        tanθ=4θ=tan-14

 



Q 2 :

A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by                                [2024]

  • sin-1[2gT2π2R]12

     

  • sin-1[π2R2gT2]12 

     

  • cos-1[2gT2π2R]12

     

  • cos-1[πR2gT2]12 

     

(1) 

       2πRT=v

       Maximum height H=v2sin2θ2g

       4R=4π2R2T22gsin2θ

       sinθ=2gT2π2Rθ=sin-1(2gT2π2R)12

 



Q 3 :

The maximum height reached by a projectile is 64 m. If the initial velocity is halved, the new maximum height of the projectile is _____ m.          [2024]



(16)     Hmax=64m

           Hmax=u2sin2θ2gH1maxH2max=u12u22

           64H2max=u2(u2)2H2max=16m

 



Q 4 :

A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at 100 m from the foot of the tower. A body of mass 2M thrown at a velocity v2from the top of the tower of height 4H will touch the ground at a distance of _____ m.            [2024]



(100)

 



Q 5 :

Projectiles A and B are thrown at angles of 45° and 60° with the vertical, respectively, from the top of a 400 m high tower. If their ranges and times of flight are the same, the ratio of their speeds of projection vA:vB is          [2024].

  • 1:3

     

  • 2:1

     

  • 1:2

     

  • 1:2

     

(4)

RA=RB

(VAcos45°)T=(VBcos60°)T

VA2=VB2

VAVB=22=12

 



Q 6 :

A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which that ball just hits step 5 of the stairway will be xms-1, where x = _____.

[Use g = 10 m/s2].            [2024]



(2)

The ball needs to just cross 4 steps to just hit the 5th step.

Therefore, horizontal range (R) = 0.4 m.

R=u·t

In vertical direction, h=12gt2

0.4=12gt2=12g(0.4u)2

u2=2u=2 m/s

Therefore, x=2

 



Q 7 :

A body starts falling freely from height H and hits an inclined plane in its path at height h. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of H/h for which the body will take the maximum time to reach the ground is ______.        [2024]



(2)

Total time of flight = T

T=2hg+2(H-h)g

For maximum time =dTdh=0

2g(-12H-h+12h)=0

H-h=h

h=H2Hh=2 



Q 8 :

A ball of mass 100 g is projected with velocity 20 m/s at 60° with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is :          [2025]

  • 20 J

     

  • 15 J

     

  • zero

     

  • 5 J

     

(2)

ki=12mv2

kf=12m(v cos 60°)2=18mv2

k=kikf=38mv2=38×0.1×400=15 J



Q 9 :

A ball having kinetic energy KE, is projected at an angle of 60° from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?          [2025]

  • KE8

     

  • KE4

     

  • KE16

     

  • KE2

     

(2)

Initial kinetic energy, K.E=12mu2

Speed at highest point, v=u cos 60°=u2

 KE2=12m(u2)2=14×12mu2=KE4



Q 10 :

Two projectiles are fired with same initial speed from same point on ground at angles of (45° – α) and (45° + α) respectively, with the horizontal direction. The ratio of their maximum heights attained is          [2025]

  • 1tan α1+tan α

     

  • 1+sin α1sin α

     

  • 1sin 2α1+sin 2α

     

  • 1+sin 2α1sin 2α

     

(3)

HMax=(u sin θ)22g

              =(Hmax)1(Hmax)2=u2sin2(45α)u2sin2(45+α)

                =(12cos α12sin α)2(12cos α+12sin α)2=1sin 2α1+sin 2α.



Q 11 :

The angle of projection of a particle is measured from the vertical axis as ϕ and the maximum height reached by the particle is hm. Here hm as function of ϕ can be represented as          [2025]

  •  

  •  

  •  

  •  

(3)

hm=u2sin2θ2g, θ=90°ϕ

 hm=u2cos2ϕ2g

From ϕ=0 to ϕ=90

hm decreases with cos2 function.



Q 12 :

A particle is projected with velocity u so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as nu225g, where value of n is : (Given 'g' is the acceleration due to gravity).           [2025]

  • 6

     

  • 18

     

  • 12

     

  • 24

     

(4)

Range R3Hmax

u2sin2θg=3u2sin2θ2g

2sinθcosθ=32sin2θ

tanθ=43  θ=53°

R=u2(2×35×45)g=24u225g



Q 13 :

Two projectiles are fired from ground with same initial speeds from same point at angles (45° + α) and (45° – α) with horizontal direction. The ratio of their times of flights is          [2025]

  • 1

     

  • 1tanα1+tanα

     

  • 1+sin2α1sin2α

     

  • 1+tanα1tanα

     

(4)

Time of flight for 1st projectile,

T1=2u sin(45+α)g

Time of flight for 2nd projectile,

T2=2u sin(45α)g

T1T2=sin(45+α)sin(45α)

T1T2=12cosα+12sinα12cosα12sinα

T1T2=cosα+sinαcosαsinα=1+tanα1tanα



Q 14 :

A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point 'O', 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is:

(Use acceleration due to gravity g = 10 m/s2 and neglect air resistance)          [2025]

  • 25 km

     

  • 4 km

     

  • 7.2 km

     

  • 22 km

     

(4)

u=360×518=100 m/s

x=u×t=100×20=2×103m=2 km

t=2Hg  H=t2g2

H=400×102=2000 m = 2 km

Displacement, D=x2+H2=22 km.



Q 15 :

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. T1 and T2 are the total flying times of first and second ball, respectively, then the ratio of T1 and T2 is :          [2025]

  • 22:1

     

  • 2 : 1

     

  • 2:1

     

  • 4:1

     

(1)

Given, (H)1=8×(H)2

H1=u2sin2θ12g and H2=u2sin2θ22g

u2sin2θ12g=8×u2sin2θ22g

sinθ1=22sinθ2  sinθ1sinθ2=22

T1=2usinθ1g and T2=2usinθ2g

 T1T2=sinθ1sinθ2=22



Q 16 :

A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1. The ratio h0h1 is ___________ [Take, g = 10 m/s2]               [2025]



(5)

Time to reach maximum height,

tHmax=usinθg=6010×12=3 s

Height in 1st second S1=30×112×10×1=25 m

Height in last second S3=30+(102)×(2×31)=5 m 

 S1S3=255=5



Q 17 :

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance up to which he can throw the same ball is        [2023]

  • 68 m

     

  • 136 m

     

  • 192 m

     

  • 272 m

     

(4)

Hmax=v22g=136 m

Rmax=v2g=2Hmax=2(136)=272 m



Q 18 :

The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of the projectile is 32u. The time of flight of the projectile is     [2023]

  • u2g

     

  • ug

     

  • 2ug

     

  • 3ug

     

(2)

ucosθ=3u2cosθ=32θ=30°

Hence T=2usin30°g=ug



Q 19 :

For a body projected at an angle with the horizontal from the ground, choose the correct statement.             [2023]

  • Gravitational potential energy is maximum at the highest point.

     

  • The horizontal component of velocity is zero at highest point.

     

  • The vertical component of momentum is maximum at the highest point.

     

  • The kinetic energy (K.E.) is zero at the highest point of projectile motion.

     

(1) 

At highest point Vy=0

Vx=ux=ucosθ

Ug=mgh, it is maximum at Hmax



Q 20 :

Two objects are projected with same velocity 'u' however at different angles α and β with the horizontal. If α+β=90°, the ratio of horizontal range of the first object to the second object will be              [2023]

  • 1 : 2

     

  • 4 : 1

     

  • 2 : 1

     

  • 1 : 1

     

(4)

Range=u2sin2θg

Range for projection angle ''α''

         R1=u2sin2αg

Range for projection angle ''β''

         R2=u2sin2βg

α+β=90° (given)  β=90°-α

R2=u2sin2(90°-α)g

R2=u2sin(180°-2α)g

R2=u2sin2αg

  R1R2=(u2sin2αg)(u2sin2αg)=11



Q 21 :

A child stands on the edge of the cliff 10 m above the ground and throws a stone horizontally with an initial speed of 5 ms-1. Neglecting the air resistance, the speed with which the stone hits the ground will be ______ ms-1 (given, g = 10 ms-2).                  [2023]

  • 20

     

  • 15

     

  • 30

     

  • 25

     

(2)

vy=2gh=200

vnet=25×200=15 m/s



Q 22 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: When a body is projected at an angle 45°, its range is maximum.

Reason R: For maximum range, the value of sin 2θ should be equal to one.

In the light of the above statements, choose the correct answer from the options given below:             [2023]

  • Both A and R are correct but R is NOT the correct explanation of A

     

  • Both A and R are correct and R is the correct explanation of A

     

  • A is false but R is true

     

  • A is true but R is false

     

(2)

R=u2gsin2θ

R is maximum for 2θ=90°



Q 23 :

Two projectiles A and B are thrown with initial velocities of 40 m/s and 60 m/s at angles 30° and 60° with the horizontal respectively. The ratio of their ranges respectively is(g=10m/s2)         [2023]

  • 1 : 1

     

  • 4 : 9

     

  • 2 : 3

     

  • 3 : 2

     

(2)

R1=u12sin2θ1g,  R2=u22sin2θ2g

R1R2=u12sin2θ1u22sin2θ2=402sin(2×30°)602sin(2×60°)=49



Q 24 :

The trajectory of a projectile, projected from the ground, is given by y=x-x220, where x and y are measured in meter. The maximum height attained by the projectile will be.             [2023]

  • 102 m

     

  • 200 m

     

  • 10 m

     

  • 5 m

     

(4)

y=x-x220

For maximum height,  dydx=01-2x20=0

x=10

So, ymax=10-10020=5 m



Q 25 :

The range of the projectile projected at an angle of 15° with the horizontal is 50 m. If the projectile is projected with the same velocity at an angle of 45° with the horizontal, then its range will be                 [2023]

  • 50 m

     

  • 502 m

     

  • 100 m

     

  • 1002 m

     

(3)

R=v2sin2θg

Rsin(2θ)

R1R2=sin(2θ1)sin(2θ2)=sin(2×15°)sin(2×45°)=sin30°sin90°

50R2=12

R2=100 m



Q 26 :

Two projectiles are projected at 30° and 60° with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is      [2023]

  • 2 : 3

     

  • 3 : 1

     

  • 1 : 3

     

  • 1 : 3

     

(3)

Hmax=u2sin2θ2g

H1H2=sin2θ1sin2θ2=13



Q 27 :

A projectile is projected at 30° from the horizontal with initial velocity 40m s-1. The velocity of the projectile at t=2 s from the start will be (Given g=10m/s2)           [2023]

  • 203 ms-1

     

  • 403 ms-1

     

  • 20 ms-1

     

  • zero

     

(1)

At t=2 particle is at maximum height moving with velocity v=40cos30°=203 ms-1



Q 28 :

Two bodies are projected from ground with same speeds 40 m s-1 at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of 60° with horizontal then sum of the maximum heights, attained by the two projectiles, is ______ m. (Given g = 10 m s-2)        [2023]



(80)

Since range is same

θ1+θ2=90°

θ2=30°

(Hmax)1+(Hmax)2=U2sin2θ12g+U2sin2θ22g

=40220(14+34)=80 m



Q 29 :

A projectile fired at 30° to the ground is observed to be at same height at time 3 s and 5 s after projection, during its flight. The speed of projection of the projectile is ______ m s-1 (Given g = 10 m s-2)              [2023]



(80)

Time of flight  t1+t2=3+5=8 sec

T=2usin30°g

8=2usin(30°)10

u=80 m/s



Q 30 :

A boy throws a ball into air at 45° from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then the value of H is _______ m.  (g=10m/s2)                                                 [2026]

  • 25

     

  • 15

     

  • 20

     

  • 10

     

(2)

T=2uyg=4

uy=402=20 m/s

y=uyΔt-12g(Δt)2

H=20×3-5×9

=60-45

=15 m