A boy throws balls into air at regular interval of $2$ second. The next ball is thrown when the velocity of first ball is zero. How high do the ball rise above his hand? [Take $g = 9.8 m / s^{2}]$ .

4.9 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.8 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

19.6 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

29.4 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $19.6 m$
Velocity of a ball becomes zero when the ball is at maximum height.
The interval is 2 s, which means that after the 1 st ball reaches max height at that time 2nd ball is thrown
It means time of ascent of ball is 2 s
So, time of ascent= u/g = 2
So u= 19.6 m/s
Then ,the max height= u^2/2g =19.619.6/29.8
Max height = 19.6 m

Suggest Corrections

Similar questions

Q. A person throws balls into air after every second. The next ball is thrown when the velocity of the first ball is zero. How high do the ball rise above his hand?

Q. A person is throwing two balls into the air, one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls at an interval of 2 seconds, how high do they rise?

[T a k e g = 10 m / s^{2}]

Q. A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is _____
(Assume,

g = 10 m s^{- 2}

)

A boy is standing on the edge of the roof at height of $78.4 m$ . He throws a ball vertically upwards with velocity of $19.6 m s^{- 1}$ . After how much time the ball will reach the ground? (Take $g = 9.8 m s^{- 2}$ )

A boy is standing on the edge of the roof at a height 78.4 m. He throws a ball vertically upwards with velocity 19.6 m/s. After how much time (in s) the ball will reach the ground? (Take a = 9.8 $m / s^{2}$ downwards)

Join BYJU'S Learning Program

A boy throws balls into air at regular interval of 2 second. The next ball is thrown when the velocity of first ball is zero. How high do the ball rise above his hand? [Take g=9.8 m/s2].

A boy throws balls into air at regular interval of $2$ second. The next ball is thrown when the velocity of first ball is zero. How high do the ball rise above his hand? [Take $g = 9.8 m / s^{2}]$ .