Q 1 :    

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t1. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t2. Time required to reach the ground, if it is dropped from the top of the tower is             [2024]

  • t1+t2

     

  • t1-t2

     

  • t2t1

     

  • t1t2

     

(3)

Using: s=ut+12at2

In situation (C): h=12gt32              ...(i)

Situation (A): -h=ut1-12gt12

h=-ut1+12gt12                        ...(ii)

Situation (B): -h=-ut2-12gt22

h=ut2+12gt22                          ...(iii)

From equation (ii) and equation (iii): eq (ii)×t2+eq (iii)×t1

ht2+ht1=12gt12t2+12gt22t1h(t2+t1)=12gt1t2(t2+t1)

h=12gt1t2                      ...(iv)

Using (i) and (iv),  12gt32=12gt1t2t3=t1·t2

 



Q 2 :    

A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is _____ m (use g = 10 ms-2).          [2024]



(45)

From A  B

-80=-2v1-12×10×22

-80=-2v1-20

-60=-2v1

v1=30 m/s

From O to A

v2=u2+2gs

302=0+2×(-10)×(-s)

900=20s

s=45 m