A person throws vertically 10 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
10m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
5m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
10cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C5cm 10 balls are thrown per second.
Time interval between two balls thrown =110s
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(110) (By first equation of motion) ⇒u=g10
Height achieved by the balls is given by v2−u2=2aS ⇒02−(g10)2=2×(−g)h ⇒h=g2×102=0.05m=5cm
The answer is (c)