A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is :

Question

A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is : (A) 45° (B) 60° (C)  tan -1 (1/2) (D)  tan -1((√3/2))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6259763ded80d00d9794dcfebb8132e3-.png" /> tan θ=( H /( R )2)=(( u 2 sin 2 45°/2 g )/(2 u 2 sin 45° cos 45°)2 g ) ( tan 45°/2)=(1/2) ⇒ θ= tan -1((1/2))

Q. A projectile is fired at an angle of $4 5^{\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is :

$4 5^{\circ}$

$6 0^{\circ}$

$tan^{- 1} \frac{1}{2}$

$tan^{- 1} (\frac{3}{2})$

$tan θ = \frac{H}{\frac{R}{2}} = \frac{\frac{u ^{2} s i n ^{2} 4 5 ^{\circ}}{2 g}}{\frac{2 u ^{2} s i n 4 5 ^{\circ} c o s 4 5 ^{\circ}}{2 g}}$
$\frac{t a n 4 5 ^{\circ}}{2} = \frac{1}{2}$
$\Rightarrow θ = tan^{- 1} (\frac{1}{2})$

Q. A projectile is fired at an angle of 45∘ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is :

45∘

60∘

tan−121​

tan−1(23​​)