Question

an object is thrown at an angle theta with horizontal if the elevation angle of the object at its highest point as seen from the point of projection is tan inverse 1 by 2 then theta is​

Answer 1

Answer:

Given:

Object thrown at angle θ with the horizontal. Angle of elevation of the object at highest point wrt Point of Projection be α = tan^-1(½)

To find:

Angle of Projection.

Concept:

Let the angle of elevation wrt Projection point be ∠α and angle of elevation wrt point of reaching the ground be ∠β

You should remember the following relationship :

$\boxed{ \sf{\tan( \theta) = \tan( \alpha ) + \tan( \beta )}}$

Seeing the symmetry of the parabola , we can say :

$\angle \alpha = \angle \beta$

So, the equation becomes :

$\tan( \theta) = \tan( \alpha ) + \tan( \alpha )$

$\implies\tan( \theta) = 2\tan( \alpha )$

Calculation:

Putting all the available values :

$\therefore\tan( \theta) = 2\tan( \alpha )$

$\implies\tan( \theta) = 2\tan \{ {tan}^{ - 1} ( \frac{1}{2}) \}$

$\implies\tan( \theta) = 2 \times \dfrac{1}{2}$

$\implies\tan( \theta) = 1$

$\implies \theta = 45 \degree$

So final answer :

$\boxed{ \red{ \huge{\theta = 45 \degree}}}$