91Ó°ÊÓ

The time taken for an object to travel of distance is t = 1.00 s .

The time of free fall and the height of fall can be determined by kinematic

equations. During the free fall, the object is fall with the gravitational

acceleration.

The kinematic equation that can be used to solve the problem is,

$∆y=v_{1}t+\frac{1}{2}a{t}^2$ (i)

Taking the origin as the point from which, the object was dropped and downward direction as the positive axis.

Notation of the terms

t = time by total distance traveled by the object.

t'= time for the object to travel from origin to

y = total distance of travel

y' = distance of object’s travel in

Now,

$$\begin{gathered} t-t\text{'}=1.00s \\ y-y\text{'}=0.50\mathrm{h} \end{gathered}$$

Suppose,$y=h$

This can be written as,

$\mathrm{h}-\mathrm{y}\text{'}=0.50\mathrm{h}$

Or

$\mathrm{y}\text{'}=0.50\mathrm{h}$

Putting the values in the equation (i)

$$\begin{gathered} y\text{'}=0.5at{\text{'}}^2 \\ t\text{'}=\sqrt{\frac{y\text{'}}{0.5a}} \\ y=0.5a{t}^2 \\ t=\sqrt{\frac{y}{0.5a}} \\ \frac{t\text{'}}{t}=\frac{\sqrt{{\displaystyle \frac{0.50h}{0.5a}}}}{\sqrt{{\displaystyle \frac{h}{0.5a}}}} \\ t\text{'}=\sqrt{0.50}×t \end{gathered}$$

From the above calculations,

$t=t-1.00s$

$\sqrt{0.50}t=t-1.00s$

$t=\frac{1}{1-\sqrt{.50}}$

$$\begin{gathered} t=\frac{1}{1Â\pm 0.707} \\ =3.41sor0.585s \end{gathered}$$

Value of t cannot be less than 1s .The correct answer is 3.41 s. So, the time of fall

of object is 3.41 s.

By using the equation

$$\begin{gathered} \mathrm{t}=\sqrt{\frac{\mathrm{h}}{0.5\mathrm{a}}} \\ \mathrm{h}=0.5×\mathrm{a}×{\mathrm{t}}^2 \\ =56.97\mathrm{m} \\ ≈57\mathrm{m} \end{gathered}$$

The height of the fall is 57 m.

The second value was lesser than 1s ,The total time cannot be less than 1s . For this

reason, we cannot use the second value.