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

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