Question

A ball is projected from top of a tower with a velocity of 5 m/s at an angle of ${53}^{\circ }$ to horizontal. Its speed when it is at a height of $0.45m$ from the point of projection is (Take $g=10m/s^2$)

• A. 2m/s
• B. 3m/s
• C. 4 m/s
• D. Data insufficient

Answer 1

The correct option is C 4 m/s

Given, projection velocity $u=5m/s$ and projection angle $\theta ={53}^{\circ }$
The horizontal component of the projection velocity $u_x=5\cos {53}^{\circ }=3m/s$
And, vertical component of the projection velocity $u_y=5\sin {53}^{\circ }=4m/s$
At the height $0.45m$ from the point of projection the vertical component of the velocity is (taking downward negative)
$v_y^2=u_y^2-2gh$
$\Rightarrow v_y^2=(4{)}^2-2\times 10\times 0.45$
$v_y=\sqrt{7}m/s$
Since, the horizontal component of velocity remain same
$\therefore$ $v_x=u_x=3m/s$
Hence, $v_{net}=\sqrt{(V_x{)}^2+(V_y{)}^2}=\sqrt{9+7}=4m/s$