Question

Two balls of equal masses are projected upward simultaneously one from the ground with speed 50m/s and other from a 40m high tower with initial speed 30m/s

Answer 1

your complete question -> Two balls of equal masses are projected upward simultaneously, one from the ground with speed 50 m/s and other from a 40 m high tower with initial speed 30 m/s. Find the maximum height attained by their centre of mass.

answer : 100 m

solution : let after time t, both ball meets at a point A.

distance covered by first ball in time t, h1 = h + u1t + 1/2 (-g)t²

= h + u1t - 1/2 gt² .......(1)

distance covered by 2nd ball in time t, h2 = u2t - 1/2 gt² ......(2)

as masses of balls are equal, so the dependence of height on time t of the centre of mass is given as,

h_c = (h1 + h2)/2

from equations (1) and (2),

h_c = h/2 + (u1 + u2)/2 t - gt²/2

it reaches maximum height when h'_c(t) = 0

so, h'_c(t) = 0 + (u1 + u2)/2 - gt

or, t = (u1 + u2)/2g

putting, u1 = 30m/s, u2 = 50m/s

so, t = (30 + 50)/2(10) = 4sec

now maximum height attained by their centre of mass, hc = h/2 + (u1 + u2)/2 t - gt²/2

= 40/2 + (30 + 50)/2 × 4 - 10(4)²/2

= 20 + 160 - 80

= 100 m

[note : here we used , g = 10m/s² . it will be change when g = 9.8 m/s² . ]

Answer 2

$\huge\bold\purple{Answer:-}$

According to the problem one ball is projected with speed of 50 m/s

and another is projected from a height of 40 m with the speed of 30 m/s

Therefore the initial velocity of the center of mass be u

let the mass of two balls are m

therefore u = m x 50 +m x 30 /2m = 40 m/s

The position of center of mass at initial time= d

d= mx 0 +m x 40 /2m = 20 m

Let the acceleration be a = -g

Therefore from the equation of kinetic energy we can say that

v^2 = u^2 +2ah

here h is the height achieved by the center of mass of the balls

0 = 40^2 - 2 x 10 x h

=>h = 80

Therefore the maximum height attained by the center of mass will be

d+h = 20+80 = 100 m