Question

please solve the question whose answer is 26.1m/s

Answer 1

after 2 sec velocity of 1st ball =V

use formula,
v = u + at
-V = 0 -9.8 × 2 = -19.6 m/s
V = 19.6 m/s

covered distance by 1st in 2sec = S
use formula,
S = ut +1/2at²
-S = 0 - 1/2 × 9.8 × 4
S = 19.6 m

e.g rest distance = 122.5 -19.6 = 102.9 m
now, 1st ball take time to cover 102.9 m distance , with initial speed = 19.6 m/s

102.9 = 19.6t +4.9t²

t² +4t -21 = 0
t² +7t -3t -21 = 0
t = -7 , and 3

but time is not negative
so, t = 3

hence, 2nd ball will reach the ground only 3sec . with speed U

122.5 = U(3) +4.9(3)²

122.5 = 3U + 4.9×9

122.5 -44.1 = 3U

78.4 = 3U

U = 78.4/3 = 26.133333 m/s

hence, initial velocity of 2nd ball = 26.1 m/s