Question

A projectile has velocity $V=(6\hat{i}+4\hat{j})m/s$ at height of $1m$ from horizontal. Then angle of projection is

• A. ${30}^o$
• B. ${37}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

Answer: (C)

${45}^o$

$V_x$ = 6 m/s and $V_y$ = 4 m/s

In Projectile Motion, Horizontal Component of velocity is constant ($U_x$ = $V_x$) = 6 m/s

For y - component,

${V_y}^2$ = ${U_y}^2$ + $2a_{y}H$

36 = ${U_y}^2$

$U_y$ = 6 m/s

Tan$\theta$ = $\frac{6}{6}$ = 1

$\theta$ = ${45}^{\circ }$