question_answer

A body is dropped from a building of height 1500 m. After two seconds, another body is dropped from the same point. How long after the first body falls, will the two bodies be 120 m apart? [\[\text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}}\]]

Answer:

Let the two bodies be 120 m apart after t seconds after the first body is dropped. Distance travelled by the1st body in ?t? sec. \[{{S}_{1}}=\frac{1}{2}g{{t}^{2}}\] Distance travelled by 2nd body in \[(t-2)s\] \[{{S}_{2}}=\frac{1}{2}g{{(t-2)}^{2}}\] Distance between the two bodies = 120 m \[\therefore \,\,{{S}_{1}}={{S}_{2}}+120m\] \[\Rightarrow \frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{(t-2)}^{2}}+120\] \[\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g({{t}^{2}}-4t+4)+120\] \[\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}-2gt+2g+120\] \[\Rightarrow -2\times 10\times t+2\times 10+120=0\] \[\Rightarrow 20t=140\Rightarrow t=\frac{140}{20}=7s\]