A bullet fired at an angle of ` 60^@` with the vertical hits the grond at a distance of `2.5 km`., Calculate the distance at which the bullet will hit the grund when fired at an angle of ` 45^@` with vertical. Assuning the speed to be the same.

Explanation

Angel or projection with horizontal,
`theta_1 = 90^@- 60^@= 30^@`
Horizontal range, ` R_1 = 2.5 km = 2500 m`
As` R_1 (u^2 sin 2 hteta)/g , so , `2500 = (u^2 sin 2 xxx 30^@)/g`
or 1 u^2/g = ( 2500)/( sin 60^@) = ( 2500)/( sqrt 3//2) = ( 2500 xx2)/( sqrt 3)`
When angle of projection is ` 45^@` with vertical, the angle or projection with horizontal,
` theta_2 = 90^2 - 45= 45^@`.
Horizontal range,
` R-2 = (u^ sin 2 xx 45^@)/g = u^@/g = ( 2500 xx2)/ (sqrt3)`
`= 2886.8 m= 2. 89 km`.