A ball is dropped from rest at lime t = 0 from certain height. A second ball is dropped from same height at time t = 1 s. At what time t, the distance between two balls becomes 10 m?
Let the first ball is dropped from rest from height h at t=0 and the second ball is dropped from same height h at t=1s. At time t, distance covered by A, H1‌‌=0⋅t+‌
1
2
gt2 ⇒H1‌‌=‌
1
2
g⋅t2‌‌(g=‌ acceleration due to gravity. ‌) At t=1s, the distance covered by the lst ball is s=‌
1
2
g(1)2 s=‌
g
2
And velocity of the ball, after covering distance s v‌‌=√2gs ‌‌=√2g‌
g
2
=g At t=1s, the 2nd ball is dropped. Let, after time t, the distance between them is 10m. The distance travelled by the 2 nd ball at time t1 is s2=‌
1
2
gt12 The total distance travelled by the 1 st ball after s1‌‌=s+vt1+‌
1
2
gt12 ‌‌=‌
g
2
+gt1+‌
1
2
gt12 s1−s2‌‌=10m ‌
g
2
+gt1+‌
1
2
gt2−‌
1
2
gt12‌‌=10 ‌
g
2
+gt1‌‌=10 gt1‌‌=10−‌
10
2
=5 t1‌‌=‌
5
g
=‌
5
10
=0.5s ∴ The time taken from rest (t=0s) by ball is ‌t=1+t1=(1+0.5)=1.5s