Given :-

• a stone is dropped from top of tower

• after 1 s , another stone is thrown vertically downward with a VELOCITY of 20 ms⁻¹

• acceleration due to gravity , g = 10 ms⁻²

To find :-

• After  travelling how long distance, second stone will overtake FIRST stone

Knowledge required :-

• second equation of MOTION

s = u t +  ½ a t²

( where s is the distance traveled , u is the initial velocity , a is the acceleration and t is the time )

Solution :-

Let, second stone overcome first stone after traveling a distance s

Now,

▸Let , time taken by stone 1 to cover distance s be t

▸initial velocity of first stone = 0

▸acceleration due to gravity , g = 10 ms⁻²

so,

Using second equation of motion

 s = u t + ½ a t²

→ s = ( 0 ) t + ½ ( 10 ) t²

→  s = ½ × 10 t²

→ s = 5 t²    ........equation (1)

Now,

▸ time taken by stone 2 to cover distance s will be ( t - 1 ) , it is because stone 2 was thrown after 1 s the first stone

▸initial velocity of second stone = 20 ms⁻¹

▸acceleration due to gravity , g = 10 ms⁻²

so,

Using second equation of motion

 s = u t + ½ a t²

→ s = (20) (t-1) + ½ (10) (t-1)²

→ s = 20 t - 20 + 5 ( t² + 1 - 2 t )

→ s = 20 t - 20 + 5 t² + 5 - 10 t

→ s = 5 t² + 10 t - 15    .......equation (2)

By equation (1) and (2)

↠   5 t² = 5 t² + 10 t - 15

↠  10 t = 15

↠  t = 15 / 10

↠  t = 1.5 s

PUTTING value of t in equation (1)

↦  s = 5 t²

↦  s = 5 ( 1.5) ²

↦  s = 5 × 2.25

↦  s = 11.25 m

Hence,

second stone will overcome first stone after traveling a distance of 11.25 metres.