question_answer

A body is thrown vertically up with a velocity of 100 m/s and another one is thrown 4 sec after the first one. How long after the first one is thrown will they meet? [Note: \[\text{g}=\text{1}0\text{m}/{{\text{s}}^{\text{2}}}\]]

Answer:

Let them meet after t sec \[{{s}_{1}}=100t-\frac{1}{2}g{{t}^{2}}\]and \[{{s}_{2}}=100(t-4)-\frac{1}{2}g{{(t-4)}^{2}}\] \[\therefore 100t-\frac{1}{2}g{{t}^{2}}=100(t-4)-\frac{1}{2}g{{(t-4)}^{2}}\] \[\therefore 400t=\frac{1}{2}g[{{t}^{2}}-(t-4)]=\frac{1}{2}g.\,\,4(2t-4)\] \[\therefore \,2t-4=\frac{800}{4g}=20,\,\,if\,\,g=10\,m/{{s}^{2}}\] \[\therefore \,\,t=12s\]