Question

From the top of a tower a ball is thrown up. It reaches the ground in $6.4sec$. A second ball thrown down with the same speed reach the ground in $3.6s$. A third ball released from the rest will reach the ground in

• A. $0.6s$
• B. $1.2s$
• C. $4.8s$
• D. $2.4s$

Answer 1

The correct option is C $4.8s$

If projection speed is $u$ then

for upward thrown ball $h=ut-g{t}^2$ or $h=u\left(6.4\right)-g(6.4{)}^2/2$ ...(1)

For downward thrown ball $h=-ut-g{t}^2$ or $h=-u\left(3.6\right)-g(\mathrm{3.6.}{)}^2/2$

or $-h=+u\left(3.6\right)+g(\mathrm{3.6.}{)}^2/2$ ....(2)

adding ...1 and ..2

we get $0=10u-14g$ or $u=1.4g$ using it in equation-1 we get value of $h=g(1.8\times 6.4)$

Suppose the free fall take time $t$ to cover $h$ then we have $h=g{t}^2/2$ use above value of $h$

then we will get $t=4.8s$