BITSAT 2015 Paper

Let the body is projected vertically upwards from A with a speed u0
Using equation, s = ut +

1

2

at2
For case (1) – h = u0t2−

1

2

gt12 ... (1)
For case (2) – h = u0t2 -

1

2

gt12 ... (2)
Subtracting eq (2) from (1), we get
0 = u0(t2+t1) +

1

2

g(t22−t12)
⇒ u0 =

1

2

g(t1−t2) ... (3)
Putting the value of u0 in eq (2), we get
- h = - (

1

2

) g (t−1−t2)t2 - (

1

2

)gt22
⇒ h =

1

2

g(t1t2) ... (4)
For case 3, u0 = 0, t = ?
- h = 0 × t - (

1

2

)gt2
h = (

1

2

)gt2
Comparing eq. (4) and (5), we get

1

2

gt2 =

1

2

gt1t2
∴ t = √t1t2