Question

At a height 0.4m from the ground, the velocity of a projectile is, $\overrightarrow{v} = (6\widehat{i} + 2\widehat{j})$ m/s. The angle of projection is: (g = 10 m/s²)

• A. 45⁰
• B. 60⁰
• C. 30⁰
• D. tan^–1(3/4)

Answer 1

Answer: (C)

30⁰

Answer 2

2 = $\sqrt { u ^ { 2 } \sin ^ { 2 } \theta - 2 h g }$

or 4 = u² sin² q – 2 (0. 4) (10)

or u² sin²q = 12

or u sin q = $2 \sqrt { 3 }$

and u cos q = 6

\ tan q = $\frac { 1 } { \sqrt { 3 } }$ or q = 30⁰