Question

a ball is thrown upward with a speed of 49m/s and at the same time another ball is dropped from the top of a tower of height 100m then after how much time they will meet each other

Answer 1

given 

ball 1 
v = 49 m/s
u= 0 m/s (as when the ball will reach its max height it will attain a speed 0 to take turn)
a=-g=-10 m/s
 
v= u+at
49= 0-10t
-4.9=t₁
as time cant be negative t =4.9

ball 2
 given
s=100m
a=g=10m/s
v=0

v² = u² +2as 
u²=v²-2as

u²=0-2*10*100
u²=-2000
therefore 
u² ≈44

a=v-u/t
at=v- u
10t=0-44
t₂=4.4
 equating the time we get
4.9-4.4=0
∴0.5=0

then both the balls will meet after 1/2 sec.

Answer 2

Answer:

A ball is thrown vertically upward with speed 49 m/s , Let after t time it strikes with another body .

then,

displacement covered by ball ( S1)

S1 = ut + 1/2at²

= 49t - 1/2 × 9.8t²

= 49t - 4.9t²

second ball dropped from the top of tower of height 100m , at same time t it strikes with first ball .

then, displacemnt covered by ball (S2) ,

S2 = 1/2gt²

-S2 = - 1/2 × 9.8t²

S2 = 4.9t²

now,

striking possible only when,

S1 + S2 = 100 m

49t - 4.9t² + 4.9t² = 100

49t = 100

t = 100/49 =2.04 sec

position of first stone from below = 49×2.04 - 4.9× (2.04)² = 79.568 m

position of second ball from the top = 4.9 × (2.04)² = 20.39 m

Explanation: