Question

A particle is thrown with a speed u at an angle Alpha with horizontal when the particle makes an angle beta with horizontal its speed will be

Answer 1

ux=ucos(alpha)
uy=usin(alpha)

when angle =beta.
vx=vcos(beta) that must be equal to ux=ucos(alpha)
so
vcos(beta)=ucos(alpha)
v=ucos(alpha)/cos(beta)

Answer 2

Given is that the particle is being thrown at angle alpha to the horizontal, therefore the component of the velocity of the particle along the two dimension will be ux along the x axis and uy along the y axis.

Further the component derived from the velocity u has the value $ux=u cos\alpha$ and uy similarly has $uy=u sin\alpha$.

If the angle changes to beta to the horizontal.

The new component of the new velocity V be Vx and Vy. As the X axis component does not change due to no change in position.

$Vx = V cos\beta = u cos\alpha$.

Therefore,

$V cos\beta = u cos\alpha$

$\Rightarrow V = u cos\alpha / cos\beta$.