question_answer

From the top of a tower of height 10 m, one fire is shot horizontally with a speed of \[5\sqrt{3}\]m/s. Another fire is shot upwards at angle of \[{{60}^{o}}\] with the horizontal at some interval of time with the same speed of \[5\sqrt{3}\] m/s. The shots collide in air at a certain point. The time interval between the two fires is

A)  2s                                          

B)  1 s    

C)  0.5 s                     

D)  0.25 s

Correct Answer: B

Solution :

                According to the questions,                 \[O-x\frac{1}{2}g{{t}^{2}}\]                 \[{{t}^{2}}=\frac{2(10-x)}{g}\]                 \[{{t}^{2}}=\frac{2(10-x)}{10}=\frac{10-x}{5}\] and        \[x=\frac{g}{2}.\,\,t\times t\]                 \[x=\frac{10}{2}{{t}^{2}}\]                 \[x=5\,{{t}^{2}}\] Putting the value of x in Eq. (i), we get                   \[{{t}^{2}}=\frac{10-x}{5}\]                 \[5{{t}^{2}}=10-x\]                 \[5{{t}^{2}}=10-5{{t}^{2}}\]                 \[10{{t}^{2}}=10\]                 \[t=1\,\,s\]