Question

An object falls a distance h from rest. If it travels 0.50h in the last 1.00 s , find

(a) the time and (b) the height of its fall. (c) Explain the physically unacceptable

solution of the quadratic equation in t that you obtain.

Short answer

a) Time of fall of the object from the original height is 3.41 s.

b) Total height of fall is 57 m.

Step-by-step solution

Given Data 

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

Understanding the concept of free fall

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)

(a) Determination of the time of fall

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)

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

From the above calculations,

$t=t-1.00s$

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

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

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

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.

(b) Determination of height of fall

By using the equation

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

The height of the fall is 57 m.

(c) Explanation of physically unacceptable solution

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

reason, we cannot use the second value.