The correct option is C 4 m/s
Given, projection velocity u=5 m/s and projection angle θ=53∘
The horizontal component of the projection velocity ux=5cos53∘=3 m/s
And, vertical component of the projection velocity uy=5sin53∘=4 m/s
At the height 0.45 m from the point of projection the vertical component of the velocity is (taking downward negative)
v2y=u2y−2gh
⇒ v2y=(4)2−2×10×0.45
vy=√7 m/s
Since, the horizontal component of velocity remain same
∴ vx=ux=3 m/s
Hence, vnet=√(Vx)2+(Vy)2=√9+7=4 m/s