A particle is projected with a velocity 40 m/s at an angle 30⁰ with the horizontal. The magnitude of change in velocity when it is at a height of 15 m from the ground is

There is no change in horizontal velocity
Vertical velocity at a height h 
$\begin{array}{l}{\mathrm{v}}_{\mathrm{y}}^2=(\mathrm{usin}\mathrm{\theta }{)}^2-2\mathrm{gy}\\ {\mathrm{v}}_{\mathrm{y}}^2={\left(40\frac{1}{2}\right)}^2-2\left(10\right)15\\ {\mathrm{v}}_{\mathrm{y}}^2=100\\ {\mathrm{v}}_{\mathrm{y}}=10\mathrm{m}/\mathrm{s}\end{array}$
Change in velocity
$\begin{array}{l}=u\sin \theta -10\\ =40\left(0.5\right)-10\\ =20-10=10 \mathrm{m}/\mathrm{s}\end{array}$