Question

A body is projected vertically upward reaches maximum height of 9.8 m/s. its time of flight is

Answer 1

Time of flight = root over 2h/g

Then h= 9.8m , g= 9.8

So, root over 2h/g = root over 2* 9.8 /9.8

Then root 2 seconds is the answer

Answer 2

Explanation:

It is given that,

The maximum height reached by the projectile, $H_{max}=9.8\ m$

The maximum height reached is given by :

$H_{max}=\dfrac{(v\ sin\theta)^2}{2g}$

$9.8=\dfrac{(v\ sin\theta)^2}{2g}$

$(v\ sin\theta)^2=2\times 9.8\times 9.8$

$v\ sin\theta=9.8\sqrt{2}$..............(1)

The time of flight of a projectile is given by :

$T=\dfrac{2v\ sin\theta}{g}$

$T=\dfrac{2\times 9.8 \sqrt{2}}{9.8}$

$T=2\sqrt2\ s$

Hence, this is the required solution.