question_answer

A particle when thrown, moves such that it passes from same height at 2 and 10 seconds, then this height h is:

A) 5g                    

B) g    

C) 8g  

D) 10g

Correct Answer: D

Solution :

          As the time taken from D to A = 2 sec. And D\[\to A\to B\to C=10\text{ sec}\left( given \right).\] As ball goes from \[B\to C\text{ }(\text{u=0, t=4sec})\] \[{{v}_{c}}=0+4g.\] As it moves from C to D, \[\text{s=ut+}\frac{1}{2}\text{a}{{\text{t}}^{2}}\] \[s=4g\times 2+\frac{1}{2}g\times 4=10g.\]