Question

A ball 'A' is dropped from the top of a tower
of height 125 m. Another ball 'B' is thrown
simultaneously with speed 20 m/s in yektically upward
direction from the base of tower. Two balls will meet
after time (g = 10 ms-2) (assume balls do not rebound
after striking ground)​

Answer 1

Answer:

see the attachment for the answer

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Ball A

u = 0 m/s.

a = + g = 10 m/s²

Let the ball A cover (125 - x) distance when it meets the second ball.

• Ball B

u = 20 m/s.

a = - g = - 10 m/s².

Let the ball B covers (x) distance before it meets the first ball.

$\huge{\bold{\underline{Explanation:-}}}$

• Ball "A".

Applying second kinematic equation.

$\bold{\large{S = ut + \frac{1}{2}at^2}}$

Substituting the values.

$\large{125 - x = 0 + \frac{1}{2} gt^2 ---(1)}$

• Ball "B".

Applying second kinematic equation.

$\bold{\large{S = ut + \frac{1}{2}at^2}}$

Substituting the values

$\large{x = 20t - \frac{1}{2}gt^2 ---(2)}$

(as body is thrown upward it acceleration due to gravity is negative.)

Add equation (1) and (2).

it comes out as

$\large{125 = 20t}$

As 1/2 gt² cancels as they are in opposite signs and x also gets cancel .

$\large{t = \frac{125}{20}}$

$\huge{\boxed{\boxed{t = 6.25 seconds}}}$

So, the body will meet after 6.25 seconds of projection.

#refer the attachment for figure.