Question

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____ s.

Answer 1

Correct Answer: 3

For case-I : 2nd equations=ut+\(\frac{1}{2}at^2\)
 H=−u(6)+\(\frac{1}{2}\)g(6)² H=−6u+18g…(i)
For case-II : h=u(1.5)+\(\frac{1}{2}\)g(1.5)²h=1.5u+\(\frac{2.25g}{2}\)…(ii)Multiplying equation (ii) by 4 we get 4h=6u+4.5 g…(iii)
equation (i) + equation (iii) 
we get 5h = 22.5gh=4.5 g… (iv) 
For case-III :h=0+\(\frac{1}{2}\)gt²…(v)Using equation (4) & equation (5) 4.5g=\(\frac{1}{2}\)gt²  t²=9⇒t=3s