Question

na Straight Line
Q.1
A ball is dropped from a high rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The
two balls meet at t=18s. What is the value of V? (take g=10ms?)
1) O
60 ms-
2) O 74 ms-1
3)
55 ms *
4)
40 ms-1​

Answer 1

Answer:

75m/s

Explanation:

the distance traveled by the first ball in 18 sec is h= (1/2)gt^2 = (1/2) * 10*18*18

=1620m

now the second ball will meet only when it covers same distance in (18-6)=12s

so for second ball

h= vt +(1/2) gt^2

1620= v*12 + (1/2) *10 * 12*12

v = (1620-720)/12

=75m/s