Question

An object is allowed to fall freely from a tower of height 39.2 m ; In exactly at the same time another stone is thrown from the bottom of the in tower in vertically upward direction with a velocity of 19.6 $ms-1$ Calculate when and where these two stones would meet ?

• A. $3s,19.6m$
• B. $2s,19.6m$
• C. $1.5s.19.6m$
• D. $4s,19.6m$

Answer 1

The correct option is B $2s,19.6m$

Distance covered by freely falling stone at any instant $t=1/2g{t}^2$

Distance covered by upward thrown stone $19.6t-\frac{1}{2}g{t}^2$

If they meet, sum=height of tower

$\begin{array}{c}\frac{1}{2}g{t}^2\&19.6t-\frac{1}{2}g{t}^2=39.9\\ 19.6t=39.2\\ t=2s\end{array}$

And they would meet at height of $19.6t1/2g{t}^2$

$=39.2-19.6=19.6m$