## Guest

Eshan
5 years ago
Dear student,

Let the height of point from ground where they meet beand time after which they meet be

Hence for first ball,

For the second ball,

Adding the two gives, t=4s, h=20m
Venkat
273 Points
5 years ago
Take height from the top as x

When dropped from top , u = 0, g = 9.8 , h = x, time taken is t
x=ut+0.5gt2 = 0 + 0.5(9.8)t2= 4.9t2   (1)

When the  ball projected upward, h = 100-x , u = 25 m/s , time is same t
100-x = 25 t-0.5*9.8*t2
x = 100-25t+4.9t2 (2)

from (1) and (2)
4.9t2 = 100-25t+4.9t2
t = 100/25 = 4 seconds

From (1)
x= 4.9(4)2    = 78.4 meters
Aman
16 Points
5 years ago
For 14
Let height reached by particle on the top of building be x
Initial velocity u= 14m/s
And final velocity v =0
As v= u+at
v= u- gt
t= 1.42s
As h= ut -1/2gt
h = 14×1.42- 1/2 ×10×1.42×1.42
= 19.88- 10.08
= 9.8m
Given total time taken= 4.6s
Time left 4.6-1.42= 3.18s
Second case::
Velocity at heighest point is zero
u=0
v =?
t=3.18s
v= 9.8×3.18= 31.164m/s
2as= v2 - u2
s= (31.164×31.164)/2×9.8
= 49.55m
Total height = 49.55m
Height of building = 49.55-9.8
= 39.7m