Question

A bullet is fired with speed `50m//s` at `45^(@)` angle with horizontal. Find the height of the bullet when its direction of motion makes angle `30^(@)` with the horizontal.

Answer 1

Correct Answer - `h=(125)/(3)m` above point of projection

`h=` height of the point where velocity makes `30^(@)` with horizontal.
As the horizontal component of velocity remain same `50 cos 45^(@)=v cos 30^(@)`
`v=50sqrt((2)/(3))`
Now by equation
`v^(2)+u^(2)2a_(y)y`
`(50xxsqrt((2)/(3)))^(2)=50^(@)-2gxh`
`rArr2gh=50^(@)-50^(2)xx(2)/(3)`
`rArr h=(2500)/(60)=(125)/(3)`
`h=(125)/(3)m` above point of projection