Question

A projectile is fired with a speed u at a n angle theta with the horizontal. When its direction
of motion makes an angle alpha with the horizontal, its speed at that instant is
(1) u cos alpha sec theta
(2) u cos theta sec alpha

(3) u cos alpha/sec theta
(4) u sec theta/cos alpha ​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

The body is thrown at a velocity "u".

Angle of projection is ${ \theta}$

$\huge{\bold{\underline{Explanation:-}}}$

As we know it is a case of projectile motion,

therefore, Velocity will have two components,

$\large{u_x = u \cos \theta}$

and,

$\large{u_y = u \sin \theta}$

As the body on travelling some distance makes an angle alpha.

At that instant be the velocity "v".

It will also have two components,

$\large{v_x = v \cos \alpha}$

and,

$\large{v_y = v \sin \alpha}$

As we know , the horizontal component of velocity does not changes.

Then,

$\large{u_x = v_x}$

Now,

$\large{u \cos \theta = v \cos \alpha}$

$\large{v = \dfrac{ u \cos \theta}{ \cos \alpha}}$

As we know,

$\large{ \dfrac{1}{ \cos \alpha} = \sec \alpha}$

Substituting the value,

$\large{v = u \cos \theta \times \sec \alpha}$

$\huge{\boxed{\boxed{v = u \cos \theta \sec \alpha}}}$

So, the the speed at that instant will be u cos(theta) sec(alpha) [Option - (2)].