A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Question

A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is (A) (v/2) √1+2 cos2 θ (B) (v/2) √1+ cos2 θ (C) (v/2) √1+3 cos2 θ (D) v cos θ

Tardigrade Admin · Accepted Answer

From figure, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a0a5157b964e1a67fdf29e85ce073e63-.png" /> average velocity, v av =(√H2+R2 / 4/T / 2)...(i) Here, H=(u2 sin 2 θ/2 g) R=(u2 sin 2 θ/g) and T=(2 u sin θ/g) Putting these value in (i), we get v av =(v/2) √1+3 cos 2 θ

Q. A particle is projected from the ground with an initial speed of v at an angle $θ$ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

$\frac{v}{2} 1 + 2 co s^{2} θ$

$\frac{v}{2} 1 + co s^{2} θ$

$\frac{v}{2} 1 + 3 co s^{2} θ$

v cos $θ$

From figure,

average velocity,
$v_{a v} = \frac{H ^{2} + R ^{2} /4}{T /2}$ ...(i)
Here, $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$
$R = \frac{u ^{2} s i n 2 θ}{g}$ and $T = \frac{2 u s i n θ}{g}$
Putting these value in (i), we get
$v_{a v} = \frac{v}{2} 1 + 3 cos^{2} θ$

Q. A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

2v​1+2cos2θ​

2v​1+cos2θ​

2v​1+3cos2θ​

v cos θ