Question

An object thrown vertically up from the ground passes the height $5m$ twice in an interval of $10s$. What is its time of flight?

• A. $\sqrt{28}s$
• B. $\sqrt{86}s$
• C. $\sqrt{104}s$
• D. $\sqrt{72}s$

Answer 1

The correct option is C $\sqrt{104}s$

We know that $\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$ $Note:$ in our sign convention upward is taken as positive

$⟹a=-g=-10m/s^2$ (as gravity acts downward)

Now,

$5=ut+\frac{1}{2}(-10)t^2$

$5=ut-5t^2$

$5t^2-ut+5=0$

It is a quadratic equation in $t$ having roots $t_1,t_2$

and difference of root $t_1-t_2=10s$ is given

We know that difference of roots can be calculated as

$t_1-t_2=\frac{\sqrt{u^2-4\times 5\times 5}}{5}=10$

$⟹\sqrt{u^2-100}=50$

$⟹u=\sqrt{2600}m/s$

We know that Time of flight $T=\frac{2u}{g}$

$⟹T=\frac{2\sqrt{2600}}{10}$

$⟹T=\sqrt{104}s$