A ball thrown up verically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c ) its position after 4 s.

Answer

intial velocity `=29.4 m s^(-1)` height =44.1 m After a s the ball will be at a distance of 4.9 m from the top or 39.2 m from the bottom .

Explanation

time of ascent is equal to the time of descent , the ball takes a total of 6 s for its upward and downward journey .
Hence , it has taken 3 s to attain the maximum height .
Final velocity of the ball at the maximum height , v=0
Acceleration due to gravity , `g=-9.8 ms^(-2)`
Equation of motion , v=u+ gt will give
`0=u+(-9.8xx3)`
`u=9.8xx3=29.4v ms^(-1)`
Hence , the ball was thrown upwards with a velovity of 29.4 `ms^(-1)`
(b) Let the maximum height attained by the ball be h .
initial velocity during the upward journey , u=29.4 `ms ^(-1)`
final velocity ,v=0
Acceleration due to gravity ,`g-=-9.8 ms ^(-2)`
From the equation of motion ,`s=ut + 1//2 at^(2)`
`h= 29.4 xx3+1//2 xx9.8xx(3)^(2)=44.1 m`

( c) Ball attains the maximum height after 3 s After attaining this height it will start falling downwards . in this case
initial velocity , u=0
Position of the after 4 s of the thrown is given by the distance travelled by it during its downward journey in 4 s -3s =1s .
Equation of motion , `s=ut +1//2 gt ^(2) `will give ,
`s=0xxt+1//2xx9.8xx1^(2)=4.9 m`
Total height =44.1 m
this means that th ball is 39.2 m (44.1 m-4.9 m) above ground after 4 seconds.