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I have the following problem to solve:

Consider a ball that is thrown with initial speed v0 at an angle a above the horizontal. If we consider it's speed at some height h above the ground show that v(h) is independent of a.

I started to solve the problem this way:

v= sqrt (vx^2 + vy^2)

vx= v0x= v0 cos a

vy= v0y- gt= v0 sin (a) -gt

Substitution of vx and vy in the first formula gives:

v= sqrt(v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2)

v^2= v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2

v^2= v0^2 ((cos^2)a + (sin^2)a) -2 v0 gt sin a + g^2 t^2

v^2= v0^2 -2 v0 gt sin a + g^2 t^2

Now I STILL have a term with a in it (-2 v0 gt sin a ) How in the world will I get rid of it?!

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# Homework Help: Projectile motion & velocity on a certain height

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