Question

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

• A. v sin q × 3
• B. v sin q + 3
• C. $\frac{v\sin\theta}{\sqrt{2}}$
• D. $\frac{v\sin\theta}{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{v\sin\theta}{\sqrt{2}}$

Answer 2

v_y² = u_y² – 2gs_y,v_y² = (usinq)² – 2g $\left( \frac { \mathrm { u } ^ { 2 } \sin ^ { 2 } \theta } { 2 \times 2 \mathrm {~g} } \right)$

̃ v_y = $\frac { v \sin \theta } { \sqrt { 2 } }$