It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation (5 π /3 6) rad should strike a given target. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target ?

Question

It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation (5 π /3 6) rad should strike a given target. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target ? (A) (5 π /3 6) textrad (B) (1 1 π /3 6) textrad (C) (7 π /3 6) textrad (D) (1 3 π /3 6) textrad

Tardigrade Admin · Accepted Answer

Ranges for complementary angles are the same Rθ = R(. (π /2) - θ .) ∴ Required angle =(π /2)-(5 π /3 6)=(1 3 π /3 6)rad

$\frac{5 π}{36} rad$

$\frac{11 π}{36} rad$

$\frac{7 π}{36} rad$

$\frac{13 π}{36} rad$

Ranges for complementary angles are the same
$R_{θ} = R_{(\frac{π}{2} - θ)}$
$∴$ Required angle $= \frac{π}{2} - \frac{5 π}{36} = \frac{13 π}{36} r a d$

Q. It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation 365π​ rad should strike a given target. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target ?

365π​rad

3611π​rad

367π​rad

3613π​rad

Ranges for complementary angles are the same Rθ​=R(2π​−θ)​ ∴ Required angle =2π​−365π​=3613π​rad

$\frac{5 π}{36} rad$

$\frac{11 π}{36} rad$

$\frac{7 π}{36} rad$

$\frac{13 π}{36} rad$

Ranges for complementary angles are the same
$R_{θ} = R_{(\frac{π}{2} - θ)}$
$∴$ Required angle $= \frac{π}{2} - \frac{5 π}{36} = \frac{13 π}{36} r a d$