A ball is projected vertically up wards with a velocity of 100 m/s. Find the speed of the ball at half the maximum height. `(g=10 m//s^(2))`

Explanation

Ships a and b are moving with same speed `20 km//h` in the direction shown in figure. It is a two dimensional, two body problem with zero acceleration. Let us find `vec(v)_(BA)`

`vec(v)_(BA)=vec(v)_(B)-vec(v)_(A)`
Here, `|vec(v)_(BA)|=sqrt((20)^(2)+(20)^(2))=20 sqrt(2) km//h`
i.e., `vec(v)_(BA)` is `20 sqrt(2) km//h` at an angle of `45^(@)` from east
towards north. Thus, the given problem can be simplified as :

A is at rest and B is moving with `vec(v)_(BA)` in the direction shown in figure.
Therefore, the minimum distance between the two is ltbtgt `s_("min")=AC=AB sin 45^(@)=10(1/sqrt(2))km=5 sqrt(2) km`
and the desired time is `t=(BC)/(|vec(v)_(BA)|)=(5sqrt(2))/(20sqrt(2))=1/4 h=15` min