question_answer

Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of 40 m/s. Then these collide at a height of (Take \[g=10\,m/{{s}^{2}}\])

A) 120 m

B) 75 m

C) 200 m

D) 45 m

Correct Answer: B

Solution :

[b] \[u=40\,m/s,\,g=10\,m/{{s}^{2}}\] Let t be time taken by the first ball to reach the highest point.          \[V=u-gt\]        \[O=40-10t\]   \[t=4s\] After reaching the first ball at the highest point now both the balls will collide after 1 sec as both the balls cover equal distances in opposite directions during 1 sec.   Therefore, the height of collision point = height gained by the second ball in 3 sec \[=40(3)-\frac{1}{2}(10){{(3)}^{2}}\] \[=120-45\] \[=75\,m\]