A projectile is fired making an angle 2θ with horizontal with velocity 4 m/s. At any instant it makes an angle θ , then its velocity is:

Question

A projectile is fired making an angle 2θ with horizontal with velocity 4 m/s. At any instant it makes an angle θ , then its velocity is: (A)  4 cos θ  (B)  4(2 cosθ - secθ )  (C)  (2 cosθ +4 secθ )  (D)  4( secθ + cosθ )

Tardigrade Admin · Accepted Answer

Horizontal component of velocity remains same.
Let

v

be the velocity, when projected with angle

θ,

then equating the horizontal velocities in both the cases, we get

v = cos θ = u cos 2 θ

\Rightarrow v = \frac{u c o s 2 θ}{c o s θ}

where

sec θ = \frac{1}{c o s θ}

∴ v = u cos 2 θ sec θ

Using

cos 2 θ = 2 cos^{2} θ - 1,

we get Given,

u = 4 m / s,

we get

v = 4 (2 cos^{2} θ - 1) sec θ

\Rightarrow v = 4 (2 cos θ - sec θ)

Q. A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m / s$ . At any instant it makes an angle $θ,$ then its velocity is:

$4 cos θ$

$4 (2 cos θ - sec θ)$

$(2 cos θ + 4 sec θ)$

$4 (sec θ + cos θ)$

Q. A projectile is fired making an angle 2θ with horizontal with velocity 4m/s. At any instant it makes an angle θ, then its velocity is:

4cosθ

4(2cosθ−secθ)

(2cosθ+4secθ)

4(secθ+cosθ)

Q. A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m / s$ . At any instant it makes an angle $θ,$ then its velocity is:

$4 cos θ$

$4 (2 cos θ - sec θ)$

$(2 cos θ + 4 sec θ)$

$4 (sec θ + cos θ)$