A fighter plane is flying horizontally at an altitude of 1.5 km with speed `720 kmh^(-1)`. At what angle of sight (w.r.t horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target? `(Take g=10ms^(-2))`

Options

A. `tan^(-1)((sqrt3)/4)`
B. `tan^(-1)((sqrt3)/2)`
C. `tan^(-1)(1/2)`
D. `tan^(-1)(2)`

Answer

A

Explanation

Here, `u=720 kmh^(-1)`
`720xx5/18 ms^(-1)=200ms^(-1)`
`H=1.5km=1.5xx1000m=1500m`
Time taken by the bomb to attack the target
`t=sqrt((2H)/g)=sqrt((2xx1500m)/(10ms^(-2)))=sqrt(300)s=10sqrt(3)s`
`R=uxxt=200ms^(-1)xx10sqrt(3)s=2000sqrt(3)m`
From fig. Angle of sight (w.r.t horizontal) `alpha=tan^(-1)(H/R)`
`alpha=tan^(-1)(1500/(2000sqrt(3))) =tan^(-1)(3/(4sqrt(3)))=tan^(-1)((sqrt(3))/4)`