Question

ball is dropped from top of tower of height h it covers a distance h/2 in the last second of its motion how long does the ball remain in air?​

Answer 1

We know that when ball is dropped u =0

$X_{nth}=u+\frac{1}{2} a(2n-1)\\\frac{h}{2} =0+(2n-1)\frac{1}{2}\times 10$

$h=10(2n-1)\\h=20n-10$

but n =1sec(last sec)

$h=20-10=10m$

$from \\S=ut+\frac{1}{2}at^{2} \\10=0+\frac{1}{2}\times 10 \times t^{2}\\ 2=t^{2}\\t=\sqrt{2} =1.414s$