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A small ball of mass m is thrown upward ...

A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force `mkv^(2)` where v is its speed. The maximum height attained by the ball is :

A

`(1)/(2k) "tan"^(-1) (k u^(2))/(g)`

B

`(1)/(k) ln (1+ (k u^(2))/(2g))`

C

`(1)/(k) "tan"^(-1) (k u^(2))/(2g)`

D

`(1)/(2k) ln (1+ (k u^(2))/(g))`

Text Solution

Verified by Experts

The correct Answer is:
D
`vec(F)= mkv^(2)- mg`
`vec(a)= (vec(F))/(m)= -[kv^(2)+g]`
`rArr v.(dv)/(dh)= -[kv^(2)+g]`
`rArr underset(u)overset(0)int (v.dv)/(kv^(2)+g)= -underset(0)overset(H)int dh`
`(1)/(2K) ln [kv^(2)+g]_(u)^(0)= -H`
`rArr (1)/(2K)ln [(ku^(2)+g)/(g)]= H`
`= (1)/(2K ) ln [1+ (ku^(2))/(g)]`
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