## anonymous 3 years ago A particle is projected with speed v at an angle alpha to the horizontal. Find the speed of the particle when it is at height h

1. anonymous

@UnkleRhaukus @ash2326

2. ash2326

@zaphod Particle's velocity has two components : 1) Horizontal Velocity= $$v\cos \alpha$$ {remains unchanged throughout the length of the motion obviously we are ignoring the air friction } 2) Vertical Velocity=$$v \sin \alpha$$ initially { decelerated by gravity} |dw:1349870216959:dw| Do you understand this part?

3. anonymous

yes now how do i get the height into this can u show the working

4. ash2326

Only $$v \sin \alpha$$ is used for vertical motion. Using $v^2-u^2=2as$ Taking upward direction to be positive We need to find vertical velocity when the particle reaches a height h $v^2-u^2=2as$ s= h a=-g $u=v\sin \alpha$ $v^2-(u\sin \alpha)^2=2\times (-g)\times h$ now find v from this This will be the vertical velocity at height h |dw:1349870813000:dw|

5. anonymous

speed will be the resultant of v cos alpha and wt u found out?

6. ash2326

Note that this v is different from the v given in the question. Let it be x $x|dw:1349871003737:dw|^2-(u\sin \alpha)^2=2\times (-g)\times h$

7. ash2326

Yeah speed will be $\sqrt {( v\ \cos \alpha)^2+x^2}$

8. ash2326

$x^2-(u\sin \alpha)^2=2\times (-g)\times h$

9. anonymous

|dw:1349871122014:dw|

10. ash2326

Actually the velocity has two components v cos alpha and x Speed will be the magnitude

11. ash2326

yes

12. anonymous

thanks :)

13. ash2326

welcome (:

14. anonymous

$speed = \sqrt{(V \sin \alpha)^{2}-2gh +(V \cos \alpha)^2}$ $= \sqrt{(V^2 (\sin ^2 \alpha +\cos^2 \alpha)-2gh}$ $= \sqrt{V^2 -2gh}$

15. anonymous

thanks algebraic :)