if u is the initial velocity of a projectile thrown by making certain angle X with the horizontal?

Question

then the instantaneous velocity at a height = (1/2)the maximum height,is u [(1-(sin^2 X)/2)]^(1/2) (i,e product of initial velocity and the square root of the difference of 1 and half the square of sine of the angle of projection). Now the question is that,as the horizontal velocity of a projectile is constant,the cosine component of this instantaneous velocity should give the horizontal velocity i,e
u cosX.But where does it ?I am unable to convert the expression of ins. velocity to u cosX.Please help

Answer 1

The formula shows v=the quantity you gave above. So the horizontal velocity=vcosx. I agree that that should be constant at any point in the flight of the projectile. If you multiply v by cosx to get vcosx, you also have to multiply the quantity you gave by cosx and then see if you can make it identical to vcosx. But I don't think you can get ucosx because then it would mean that vcosx=ucosx which is not correct.

Answer 2

If there is no change in the internal energy of the projectile, the final velocity would be equal to the initial velocity. This means the final kinetic energy would be equal to the initial kinetic energy. The C is equal to the difference of the initial and final kinetic energies. Initial KE = ½ * 5 * 200^2, Final KE = ½ * 5 * 150^2 Decrease of internal energy = ½ * 5 * 200^2 – ½ * 5 * 150^2 = 43,750 J

Answer 3

Your'e confusing the initial angle with the instantaneous angle. The vertical velocity component is constantly changing, so the instantaneous angle is constantly changin.

Just consider that at the top of the arc, the angle must be 0º, since there is vertical velocity is zero at the top.