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)

     

(C)  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 

     

(A)  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)