Question

A ball is dropped from height of 20m at the same instant and the ball is thrown up from the ground with speed 20m/s when and where will ball meet​

Answer 1

Question : A ball is dropped from height of 20m at the same instant and the ball is thrown up from the ground with speed 20m/s when and where will ball meet.

Initial velocity of the vall = 0 ( Let's take it as U)

Acceleration = 10 ms^-2 (Let's take it as A)

Distance = 20m (Let's take it as S)

From the equation of motion =

v²-u² = 2as

v² - 0 = 2 × 10 × 20

v² - 400

v= 20 ms^-1

From the first equation of motion,

v = u + at

20 = 0 + 10 × T

T = 2 seconds.

Answer 2

Solution:

Let us assume that the balls meet at a height "h" after time "t" above the ground.

For the ball dropped from the height:

Distance covered by ball = (20 - h)m.

=> u = 0 m/s

=> s = (20 - h)m

=> g = 10 m/s²

By using 2nd equation of motion.

s = ut + 1/2 at²

20 - h = 0 + 1/2×10t²

20 - h = 5t²          ...........(1)

For the ball thrown vertically upwards:

=> u = 20 m/s

=> s = h

=> g = -10 m/s²

By using 2nd equation of motion:

h = ut + 1/2at²

h = 20t - 5t²           ............(2)

Adding eq(1) and eq(2), we get

=> 20 - h + h = 5t² + 20t - 5t²

=> 20 = 20t

=> t = 1s

Put the value of t in eq (1)

=> 20 - h = 5t²

=> 20 - h = 5

=> h = 15 m

So, two balls meet at a height 15m from the ground after 1 s.