Question

Salim is standing on the terrace of a building. He throws a ball vertically upwards. The ball covers a distance $1\frac{3}{4}m$and then starts falling down. If it falls $2\frac{2}{3}m$vertically downward. Where is the ball now from the Salim?

Answer 1

Given,

Distance covered vertically upward$=1\frac{3}{4}m$

and distance traveled vertically downward$=2\frac{2}{3}m$

Let us denote the distance traveled towards vertical upward by a positive sign. So the distance traveled vertically downward would be denoted by a negative sign.

Thus, the distance traveled by a ball from Salim would be

$$\begin{gathered} 1\frac{3}{4}+\left(-2\frac{2}{3}\right)=\frac{7}{4}+\left(\frac{-8}{3}\right) \\ =\frac{7\times 3}{4\times 3}+\frac{-8\times 4}{3\times 4}\left\{LCM\right(3,4)=12\} \\ =\frac{21}{12}+\frac{-32}{12} \\ =\frac{21+(-32)}{12} \\ =\frac{-11}{12}m \end{gathered}$$

The negative shows the ball is in the downward direction from Salim.

Hence, the ball is $\frac{11}{12}m$ vertically below from the Salim.