Question

Two stones are thrown vertically upwards with the same velocity of 49m/s.if they are thrown one after the other with a time lapse of 3sec height at which they collide is

Answer 1

$$\begin{gathered} Dis\tan cetravelledby1ststone, \\ S=ut+\frac{1}{2}a{t}^2 \\ x=49\times t-\frac{1}{2}\times 9.8\times t^2.......\left(1\right) \\ Forsecondstone, \\ x=49\times \left(t-2\right)-\frac{1}{2}\times 9.8\times {\left(t-2\right)}^2............\left(2\right) \\ Onequating\left(1\right)\&\left(2\right) \\ 49\times t-\frac{1}{2}\times 9.8\times t^2=49\times \left(t-2\right)-\frac{1}{2}\times 9.8\times {\left(t-2\right)}^2 \\ 49t-4.9t^2=49t-98-4.9\left(t^2+4-4t\right) \\ -4.9t^2+98=-4.9t^2-19.6+19.6t \\ \frac{98+19.6}{19.6}=t \\ t=6sec \\ So,theywillmeetafter6sec. \\ x=49\times 6-\frac{1}{2}\times 9.8\times 6^2 \\ =294-176.4=117.6m \\ So,theywillcollideatheight=117.6mfromground. \end{gathered}$$