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  • A ball is dropped from the top of a building 100m high. At the same time, another ball is thrown upwards with a velocity of 40 m/sec from the bottom of the building. The two balls will meet after T

A ball is dropped from the top of a building 100m high. At the same time, another ball is thrown upwards with a velocity of 40 m/sec from the bottom of the building. The two balls will meet after T seconds. The value of 10T will be -

Option: 1

25


Option: 2

20


Option: 3

30


Option: 4

50


Answers (1)

best_answer

         

 

\\ s_{1}+s_{2}=H=100 m \\ s_{1}= ut+\frac{1}{2}at^{2} \\\\ s_{1} =0+\frac{1}{2}gt^{2} \\ \\ s_{2}=ut+\frac{1}{2}at^{2} \\ \\ s_{2}=40t-\frac{1}{2}gt^{2}\\ \\ \text{Now ,} \\ 100=\frac{1}{2}gt^{2}+40t-\frac{1}{2}gt^{2} \\ \\ 100=40t \\\\t=\frac{100}{40}=2.5 s

\therefore 10t=10\times 2.5=25 s

Posted by

Pankaj Sanodiya

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