A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is

Question

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is (A) v sin θ × 3 (B) (v sin θ/3) (C) (v sin θ/√2) (D) (v sin θ/√3)

Tardigrade Admin · Accepted Answer

vB2=v2 sin 2 θ-(2 g/2)((u2 sin 2 θ/2 g)) vB2=(v2 sin 2 θ/2) vB=(v sin θ/√2)

Q. A projectile is thrown with velocity $v$ at an angle $θ$ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is

$v sin θ \times 3$

$\frac{v s i n θ}{3}$

$\frac{v s i n θ}{2}$

$\frac{v s i n θ}{3}$

$v_{B}^{2} = v^{2} sin^{2} θ - \frac{2 g}{2} (\frac{u ^{2} s i n ^{2} θ}{2 g})$
$v_{B}^{2} = \frac{v ^{2} s i n ^{2} θ}{2}$
$v_{B} = \frac{v s i n θ}{2}$

Q. A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is

vsinθ×3

3vsinθ​

2​vsinθ​

3​vsinθ​

vB2​=v2sin2θ−22g​(2gu2sin2θ​) vB2​=2v2sin2θ​ vB​=2​vsinθ​

Q. A projectile is thrown with velocity $v$ at an angle $θ$ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is

$v sin θ \times 3$

$\frac{v s i n θ}{3}$

$\frac{v s i n θ}{2}$

$\frac{v s i n θ}{3}$

$v_{B}^{2} = v^{2} sin^{2} θ - \frac{2 g}{2} (\frac{u ^{2} s i n ^{2} θ}{2 g})$
$v_{B}^{2} = \frac{v ^{2} s i n ^{2} θ}{2}$
$v_{B} = \frac{v s i n θ}{2}$