A person throws vertically n balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is
A
gn2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2gn
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
g2n2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2gn2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cg2n2 n balls are thrown per second. Time interval between two balls thrown =1nsec This is the time that a ball takes to reach its highest point Using first equation of motion, v=u+at As the velocity of ball is 0 at its maximum height ⇒0=u−g(1n) ⇒u=gn Height achieved by the balls is given by v2−u2=2as ⇒02−(gn)2=2×(−g)h ⇒h=g2n2