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

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Solution

The key here is that, for a projectile (defined to be moving only under the influence of gravitational acceleration), there is no acceleration in the horizontal direction. Hence, its horizontal speed is constant during its entire flight (from the moment it leaves the launcher until it hits something, and is no longer a projectile).

Horizontal velocity remains same = u cos(alpha)

Let particle velocity at time when it makes angle beta be v m/s.

Then,

Vertical velocity = v cos(beta)

horizontal velocity = vertical velocity
u cos(alpha) = v cos(beta)
v = u cos(alpha)/cos(beta)
=> v=ucos(alpha)sec(beta)

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