A particle is thrown with the speed u at angle(A) with the horizontal. When the particle makes an angle (B) with horizontal then what will be its speed

Question

Motion in plane questions

Answer 1

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).  

 

Given that it is thrown (launched) with speed u at angle A, we can calculate the horizontal component of its initial (launch) speed, v_x, thinking of the situation as a right triangle with overall speed u on the hypotenuse, and v_x being the adjacent side.  Thus:

 

v_x = u*cos(A)

 

If you cannot figure out to determine this trigonometrically, I can help you with that in more detail (though without a diagram, it may be harder to follow).

 

At the later time, the projectile still has this horizontal component to its velocity, v_x (since the horizontal speed of a projectile is constant).  If we know that, and are given the new angle B, solving for the new overall speed at that time means solving for the hypotenuse of a right triangle where v_x (=u*cos(A)) is the side adjacent to angle B.  If you can do that, you can determine the overall speed of the projectile at angle B in terms of given quantities (B, A, u).

 

If you would like to go into more detail about the algebra and trigonometry of the ultimate solution, I would be happy to do that.

 

I hope this helps!