Question

Maximum height of 9.8m. Its flight is
$\begin{aligned} & \left( 1 \right)4s \\\ & \left( 2 \right)2s \\\ & \left( 3 \right)\sqrt{3}s \\\ & (4)\sqrt{2}s \\\ \end{aligned}$

Answer 1

First using the equation for maximum height, calculate the initial velocity. Then calculate the time of flight. We know that the time of flight is the ratio of initial velocity to the acceleration due to gravity. Thus by substituting both the initial velocity and the acceleration due to gravity we will get the time of flight.
Formula used:
Maximum height,
$H=\dfrac{{{u}^{2}}}{2g}$
Where, u is the initial velocity
g is the acceleration due to gravity.
H is the maximum height.
Time of flight ,
$T=\dfrac{u}{g}$
where, u is the initial velocity.
g is the acceleration due to gravity.

Complete answer:
Given that maximum height,
H=9.8m
Then,
$H=\dfrac{{{u}^{2}}}{2g}$ ……..(1)
where, u is the initial velocity
g is the acceleration due to gravity.
To find initial velocity rearrange equation (1),
$\Rightarrow {{u}^{2}}=2gH$
On rearranging we get the equation of the square of initial velocity.
Then by substituting the values we get,
${{u}^{2}}=2\times 9.8\times 9.8$
$\Rightarrow {{u}^{2}}=2\times {{9.8}^{2}}$
Thus on taking the square root of the above value we will get the value of initial velocity.
That is,
$\Rightarrow u=9.8\sqrt{2}$ $m/s$
Then the time of flight is given by,
$T=\dfrac{u}{g}$
Substitute the value of initial velocity and acceleration due to gravity we get the time of flight,
$\Rightarrow T=\dfrac{9.8\sqrt{2}}{9.8}$
$\Rightarrow T=\sqrt{2}s$

So, the correct answer is “Option D”.

Note:
Here, a projectile motion is shown. An object that is in flight after being thrown or projected is called projectile. Such projectiles can be any object. The time of maximum height is the time taken to reach the maximum height. These equations can only be used in the cases of projectile motions.