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=4⇒θ=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=2gT2π2R⇒θ=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=u2sin2θ2g⇒H1maxH2max=u12u22

           ⇒64H2max=u2(u2)2⇒H2max=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=2⇒u=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=H2⇒Hh=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=ki–kf=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]

  • 1–tan α1+tan α

     

  • 1+sin α1–sin α

     

  • 1–sin 2α1+sin 2α

     

  • 1+sin 2α1–sin 2α

     

(3)

HMax=(u sin θ)22g

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

                =(12cos α–12sin α)2(12cos α+12sin α)2=1–sin 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 R = 3Hmax

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

     

  • 1–tanα1+tanα

     

  • 1+sin2α1–sin2α

     

  • 1+tanα1–tanα

     

(4)

Time of flight for 1st projectile,

T1=2u sin(45+α)g

Time of flight for 2nd projectile,

T2=2u sin(45–α)g

T1T2=sin(45+α)sin(45–α)

T1T2=12cosα+12sinα12cosα–12sinα

T1T2=cosα+sinαcosα–sinα=1+tanα1–tanα



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 h0 : h1 is ___________ [Take, g = 10 m/s2]               [2025]



(5)

Time to reach maximum height,

tHmax=usinθg=6010×12=3 s

Height in 1st second S1=30×1–12×10×1=25 m

Height in last second S3=30+(–102)×(2×3–1)=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=3u2⇒cosθ=32⇒θ=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=u2g sin2θ

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=10 m/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=0→1-2x20=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

R∝sin(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