Home
Class 11
PHYSICS
A man throws balls with the same speed v...

A man throws balls with the same speed vertically upwards one after the other at an interval of 2s. What should be the speed of the throw so that more than two balls are in the sky at any time ? (Given `g = 9.8 m//s^(2)`)

A

1) `("Any speed less than")19.6ms^(-1)`

B

2) `("Only with speed")19.6ms^(-3)`

C

3) `("More than")19.6 ms^(-4)`

D

4) `("At least")9.8ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Time taken by ball to reach maximum height
`v=u-gt`
at meximum height , final speed is Zereo ie, v=0
So , u=gt
or t=u.g
In 2s. U=`2xx9.8=196ms^(-1)`
If man throws the ball with velocity of `19.6ms6(-1)`
Then after 2 s it will reach the maximum height . When he thrown 2nd ball , 1st is at top. when he throws third ball , 1st will come to ground and 2nd will at the top. Therefore , only 2balls are in air . if he wants to keep more than 2 balls in air he should throw the ball with a speed greater than `19.6ms^(-1)`
Promotional Banner

Similar Questions

Explore conceptually related problems

A man throws ball with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two ball are in the sky at any time (Given g=10(m)/(2^2) )

A person throws balls into the air one after the other at an interval ofone second. The next ball is thrown when the velocityof the ball thrown earlier is zero. To what height the ball rise:

A boy is standing on top of a tower of height 85 m and throws a ball in the vertically upward direction with a certain speed. If 5.25 seconds later he hears the ball hitting the ground, then the speed with which the boy threw the ball is (take g = 10 m/s^(2) , speed of sound in air = 340 m/s)

Two balls are released from the same height at an interval of 2s. When will the separation between the balls be 20m after the first ball is released? Take g=10m//s^(2) .

Two balls are projected vertically upward with same speed 35 m/s in 3s interval. Find height at which both balls collide.

A man throws a ball weighing 500 g vertically upwards with a speed of 10 m/s (i) What will be its initial momentum? (ii) What would be its momentum at the highest point of its flight?

Two balls are projected simultaneously with the same speed from the top of a tower, one vertically upwards and the other vertically downwards. They reach the ground in 9 s and 4 s, respectively. The height of the tower is (g=10 m//s^(2))

A ball is projected vertically up wards with a velocity of 100 m/s. Find the speed of the ball at half the maximum height. (g=10 m//s^(2))

A ball thrown vertically upwards with an initial velocity of 1.4 m/s returns in 2s. The total dispalcement of the ball is