A ball thrown in vertically upward direction attains maximum height of 16 m. At what height would its velocity be half of its initial velocity ?

A ball thrown in vertically upward direction attains maximum height of

1.	A ball thrown in vertically upward direction attains maximum height of 16 m. At what height would its velocity be half of its initial velocity ?
Answer» Solution :For motion of a ball in vertically upward direction, Suppose `v _(0) =` initial velocity, v = velocity at MAX. HEIGHT (of 16 m) `=0, a =-g, d = 16 m` `THEREFORE` From equation` 2ad = v ^(2) - v _(0) ^(2) ` we have `= 2g XX 16 = (0) ^(2) -v_(0) ^(2)` `therefore -32 g =-v _(0) ^(2)` ` therefore v _(0) ^(2) - 32 g ...(i)` Now suppose velocity at height `h = v = (v_(0))/(2)` From the equation `2 ad = v ^(2) - v _(0) ^(2) ` we get, `-2gh = (v_(0) ^(2))/(4) - v _(0) ^(*2),` where `a =- g , d = h , v = (v _(0))/(2)` `therefore - 2gh = (-3 v _(0) ^(2))/(4)` `therefore h = (3v _(0) ^(2))/(4 xx 2g)` `therefore h = ( 3 xx 32 g )/(8g) ` (From EQ. (i)) =12 m.