• question_answer 29) At time t = 0. A bullet is fired vertically upwards with-a speed of$98\text{ }m{{s}^{-1}}.$ A second bullet is fired vertically upwards with the same speed at $t=5$s. Then: A)  the two bullets will be at the same height above the ground at t = 12.5 s.                  B)  the two bullets will reach back their stating  point at t = 25 s.                           C)  the two bullets will have the same speed at  t = 20 s.                                 D)  the maximum height attained by either bullet will be 980 m.

The two bullets will be at the same height at $t=n$if $98-4.9\,{{n}^{2}}=98(n-5)-4.9\,{{(n-5)}^{2}}.$ This gives$~n=12.5\text{ }s.$The time to reach the highest point is given by 0 = 98 - 9.8 t, i.e., t = 10 sec. Therefore, the time of flight is $2\times 10\,s$ or 20 sec. The speed of the first bullet at t = 20 sec. is 98 m/s again while that of the second bullet is $(98-9.8\times 0.5)\,m/s$or 49 m/s. The maximum heights attained are 490 m each. Hence, the correction option is [a].