Question

Two balls X and Y are thrown from top of tower one
vertically upward and other vertically downward with
same speed. If times taken by them to reach
the ground are 6 s and 2 s respectively, then the
height of the tower and initial speed of each ball are
(g = 10 m/s)
(1) 60 m, 15 m/s (2) 80 m, 20 m/s
(3) 60 m, 20 m/s (4) 45 m, 10 m/s
leration​

Answer 1

Two balls X and Y are thrown from top of tower one
vertically upward and other vertically downward with
same speed. If times taken by them to reach
the ground are 6 s and 2 s respectively, then the
height of the tower and initial speed of each ball are
(g = 10 m/s)
(1) 60 m, 15 m/s (2) 80 m, 20 m/s
(3) 60 m, 20 m/s (4) 45 m, 10 m/s
leration​

Answer 2

Answer:

Given : 

Case 1:

For the ball A :

initialSpeed=u=20m/s[upwards]=-20m/s

a=10m/s²[downwards]=+10m/s²

time=t=tsec

Distance=Sa meters

From seond equation of motion,

Sa=ut+1/2at²

Sa=-20t+1/2x10xt²

Sa=-20t+5t² --------------equation(1)

Case II:

For ball B:

Initial speed=u=20m/s[downwards]=+20m/s

a=10m/s²[downwards]=+10m/s²

time =t sec

Distance =Sb meters.=(40-sa)

From seond equation of motion,

Sb=ut+1/2at²(4-Sa)

=20t+5t² ====[equation2]

Therefore , from equation 1 and 2 we get:

40=-20t+5t²+20t+5t²40

=10t²4=t²

t=2sec

Let us find out distance Sa in 2 sec

Sa=ut+1/2at²

sa=-40+20=-20m

From the ground that is at 20m the two balls will collide.