A ball thrown up vertically returns to the thrower after 6s.
(a) The velocity with which it was thrown up,
The maximum height it reaches, and
© Its position after 4s

Explanation

Time ascent is equal to the time of descent. The ball takes a total of 6s for its upward and downward journey.
Hence, it has taken 3s to attain the maximum height.
Final velocity of the ball at tha maximum height, v=o. Acceleration due to gravity, `g=-9.8ms^(-2)`
Equation of motion, v=u+gt will give
`o=u+(-9.8xx3)`
`u=9.8xx3=29.4ms^(-1)`
Hence, the ball was thrown upwards with a velocity of `29.4ms^(-1)`
Let the maximum height attained by the ball be .h. Initial velocity during the upward journey `u=29.4ms^(-1)`
Final velocity, v=o
Acceleration due to gravity, `g=-9.8ms^(-2)` From the equation of motion,
`s=ut+1//2at^(2)`
`h=29.4xx3+1//2xx-9.8xx(3)^(2)=44.1m`
Ball attains the maximum height after 3s. After attaining this height will start falling downwards.
In this case, Initial velocity u=o
Position of the ball after 4s of the throw is given by the distance travelled by it during its downward journey is `4s-3s=1s`.
Equation of motion, `s-ut+1//2g""t^(2)` will give.
`S=Oxt+1//2xx9.8xx12=4.9m`
Total height =44.1m
This means that the ball is 39.2m
(44.1m -4.9m) above the ground after 4 seconds.