A person, standing on the roof of a 40 m high tower, throws a ball vertically upwards with speed 10 m/s. Two seconds later, he throws another ball again in vertical direction (use `g=10(m)/(s^2)`). Both the balls hit the ground simultaneously.

Options

A. The first ball hits the ground after 4 seconds.
B. The second ball was projected vertically downwards with speed 5 m/s
C. The second ball was projected vertically downwards with speed 5 m/s.
D. the distance travelled by the first ball is 10 m greater than the distance travelled by the second ball.

Answer

A, C, D

Explanation

For Ball A
`S=-40m`,`u=+10(m)/(s)`,
`a=-10(m)/(s^2)`
`S=ut+(1)/(2)at^2-40=10t+(1)/(2)(-10)t^2`
`impliest=-2`,`4impliest=4` sec
(Ball a will reach ground after 4 sec)
In two seconds, displacement of Ball A is
`S(2)=10t-(1)/(2)(10)t^2=10(2)-(4(10))/(2)=0`
i.e. The ball B is thrown when the first ball have come back to the initial position, having velocity 10 m/s in opposite direction (downward direction).
For both the balls to reach ground simultaneously initial velocity in y-direction must be same for both the balls.
Thus the second ball was projected vertically downwards with speed of 10 m/s.
Distance travelled by first ball (a) in two seconds
`=2xx[10xx1-(1)/(2)xx10xx1^2]=2xx5=10`m.
After that ball A and Ball B will travel same distance (`=40m`). Since both balls have same velocity and acceleration at `t=2` sec, hence they will hit the ground with same velocity.