2.9 Vertical motion under gravity (application)  (Page 2/2)

Problem : A particle is released from a height of “3h”. Find the ratios of time taken to fall through equal heights “h”.

Solution : Let $t_1$ , $t_2$ and $t_3$ be the time taken to fall through successive heights “h”. Then, the vertical linear distances traveled are :

$$\begin{array}{l}h=\frac{1}{2}g{t_1}^2\\ \mathrm{2h}=\frac{1}{2}g{(t_1+t_2)}^2\\ \mathrm{3h}=\frac{1}{2}g{(t_1+t_2+t_3)}^2\end{array}$$

$$\begin{array}{l}\Rightarrow {t_1}^2:{(t_1+t_2)}^2:{(t_1+t_2+t_3)}^2::1:2:3\end{array}$$

$$\begin{array}{l}\Rightarrow t_1:t_1+t_2:t_1+t_2+t_3::1:\surd 2:\surd 3\end{array}$$

In order to reduce this proportion into the one as required in the question, we carry out a general reduction of the similar type. In the end, we shall apply the result to the above proportion. Now, a general reduction of the proportion of this type is carried out as below. Let :

$$\begin{array}{l}{t}_1:t_1+t_2:t_1+t_2+t_3::a:b:c\end{array}$$

$$\begin{array}{l}\Rightarrow \frac{t_1}{a}=\frac{t_1+t_2}{b}=\frac{t_1+t_2+t_3}{c}\end{array}$$

From the first two ratios,

$$\begin{array}{l}\Rightarrow \frac{t_1}{a}=\frac{t_1+t_2-t_1}{b-a}=\frac{t_2}{b-a}\end{array}$$

Similarly, from the last two ratios,

$$\begin{array}{l}\Rightarrow \frac{t_1+t_2}{b}=\frac{t_1+t_2+t_3-t_1-t_2}{c-b}=\frac{t_3}{c-b}\end{array}$$

Now, combining the reduced ratios, we have :

$$\begin{array}{l}\Rightarrow \frac{t_1}{a}=\frac{t_2}{b-a}=\frac{t_3}{c-b}\end{array}$$

$$\begin{array}{l}\Rightarrow t_1:t_2:t_3::a:b-a:c-b\end{array}$$

In the nutshell, we conclude that if

$$\begin{array}{l}{t}_1:t_1+t_2:t_1+t_2+t_3::a:b:c\end{array}$$

Then,

$$\begin{array}{l}{t}_1:t_2:t_3::a:b-a:c-b\end{array}$$

Applying this result as obtained to the question in hand,

$$\begin{array}{l}{t}_1:t_1+t_2:t_1+t_2+t_3::1:\surd 2:\surd 3\end{array}$$

We have :

$$\begin{array}{l}\Rightarrow t_1:t_2:t_3::1:\surd 2-1:\surd 3-\surd 2\end{array}$$

Problem : Balls are successively dropped one after another at an equal interval from a tower. At the instant, $9^{\mathrm{th}}$ ball is released, the first ball hits the ground. Which of the ball in series is at ¾ of the height of the tower?

Solution : Let “t” be the equal time interval. The first ball hits the ground when $9^{\mathrm{th}}$ ball is dropped. It means that the first ball has fallen for a total time (9-1)t = 8t. Let $n^{\mathrm{th}}$ ball is at 3/4th of the height of tower. Then,

For the first ball,

$$\begin{array}{l}\Rightarrow h=\frac{1}{2}gx{8t}^2\end{array}$$

For the nth ball,

$$\begin{array}{l}\Rightarrow h-\frac{3h}{4}=\frac{h}{4}=\frac{1}{2}gx{(9-n)}^2{t}^2\end{array}$$

Combining two equations, we have :

$$\begin{array}{l}\Rightarrow h=2gx{(9-n)}^2{t}^2=\frac{1}{2}gx{8t}^2=32g{t}^2\\ \Rightarrow {(9-t)}^2=16\\ \Rightarrow 9-n=4\\ \Rightarrow n=5\end{array}$$

Problem : A ball is dropped vertically from a tower. If the vertical distance covered in the last second is equal to the distance covered in first 3 seconds, then find the height of the tower. Consider g = 10 $m/s^2$ .

Solution : Let us consider that the ball covers the height $y_n$ in $n^{\mathrm{th}}$ second. Then, the distance covered in the $n^{\mathrm{th}}$ second is given as :

$$\begin{array}{l}{y}_n=u+\frac{g}{2}x(2n-1)=0+\frac{10}{2}x(2n-1)=10n-5\end{array}$$

On the other hand, the distance covered in first 3 seconds is :

$$\begin{array}{l}{y}_3=ut+\frac{1}{2}g{t}^2=0+\frac{1}{2}x10x{3}^2=45m\end{array}$$

According to question,

$$\begin{array}{l}{y}_n=y_3\end{array}$$

$$\begin{array}{l}\Rightarrow 10n-5=45\\ \Rightarrow n=5s\end{array}$$

Therefre, the height of the tower is :

$$\begin{array}{l}\Rightarrow y=\frac{1}{2}x10x{5}^2=125m\end{array}$$

Problem : A ball, thrown vertically upward from the ground, crosses a point “A” in time “ $t_1$ ”. If the ball continues to move up and then return to the ground in additional time “ $t_2$ ”, then find the height of “A” from the ground. .

Solution : Let the upward direction be the positive reference direction. The displacement equation for vertical projection from the ground to the point “A”, is :

$$\begin{array}{l}h=u{t}_1\mathrm{+}\frac{1}{2}g{t_1}^2\end{array}$$

In this equation, the only variable that we do not know is initial velocity. In order to determine initial velocity, we consider the complete upward motion till the ball reaches the maximum height. We know that the ball takes half of the total time to reach maximum height. It means that time for upward motion till maximum height is :

$$\begin{array}{l}{t}_{\frac{1}{2}}=\frac{t_1\mathrm{+}{t}_2}{2}\end{array}$$

As we know the time of flight, final velocity and acceleration, we can know initial velocity :

$$\begin{array}{l}0=u-g{t}_{\frac{1}{2}}\\ \Rightarrow u=g{t}_{\frac{1}{2}}=\frac{\left(t_1\mathrm{+}{t}_2\right)g}{2}\end{array}$$

Substituting this value, the height of point “A” is :

$$\begin{array}{l}h=u{t}_1\mathrm{+}\frac{1}{2}g{t_1}^2\\ \Rightarrow h=\frac{\left(t_1\mathrm{+}{t}_2\right)g}{2}x{t}_1\mathrm{-}\frac{1}{2}g{t_1}^2\\ \Rightarrow h=\frac{g{t_1}^2+g{t}_1{t}_2-g{t_1}^2}{2}\\ \Rightarrow h=\frac{g{t}_1{t}_2}{2}\end{array}$$

Problem : A ball, thrown vertically upward from the ground, crosses a point “A” at a height 80 meters from the ground. If the ball returns to the same position after 6 second, then find the velocity of projection. (Consider g = 10 $m/s^2$ ).

Solution : Since two time instants for the same position is given, it is indicative that we may use the displacement equation as it may turn out to be quadratic equation in time “t”. The displacement, “y”, is (considering upward direction as positive y-direction) :

$$\begin{array}{l}y=ut\mathrm{+}\frac{1}{2}g{t}^2\\ 80=ut\mathrm{+}\frac{1}{2}x\mathrm{-}10t^2=ut\mathrm{-}5t^2\\ 5t^2-ut\mathrm{+}80\mathrm{=}0\end{array}$$

We can express time “t”, using quadratic formulae,

$$\begin{array}{l}\Rightarrow t_2=\frac{-(-u)\mathrm{+}\surd \{{(-u)}^2-4x5x80\}}{2x5}=\frac{u\mathrm{+}\surd (u^2-1600)}{10}\end{array}$$

Similarly,

$$\begin{array}{l}\Rightarrow t_1=\frac{u\mathrm{-}\surd (u^2-1600)}{10}\end{array}$$

According to question,

$$\begin{array}{l}\Rightarrow t_2\mathrm{-}{t}_1=\frac{\surd (u^2-1600)}{5}\\ \Rightarrow 6=\frac{\surd (u^2-1600)}{5}\\ \Rightarrow \surd (u^2-1600)=30\end{array}$$

Squaring both sides we have,

$$\begin{array}{l}\Rightarrow u^2-1600=900\\ \Rightarrow u=50m/s\end{array}$$