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 $\sqrt{104}s$

The object has flight height $(s-5)$ m in $\frac{10}{2}s$= 5s

The object is descending at the same time= 5s

Using,

$s=ut+\frac{1}{2}a{t}^2$

$s-5=0\times 5+\frac{1}{2}\times 9.8\times 5^2$

s=127.5 m

Time taken to travel vertically 127.5 m,

$=>127.5=0\times t+\frac{1}{2}\times 9.8\times t^2$

$t^2=26.02s$

t= 5.1 s

Therefore total time of flight is= $2\times 5.1$ = 10.2s

$\sqrt{104}s$