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zaphod
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
@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?
yes now how do i get the height into this can u show the working
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|
speed will be the resultant of v cos alpha and wt u found out?
Note that this v is different from the v given in the question. Let it be x \[x|dw:1349871003737:dw|^2-(u\sin \alpha)^2=2\times (-g)\times h\]
Yeah speed will be \[\sqrt {( v\ \cos \alpha)^2+x^2}\]
\[x^2-(u\sin \alpha)^2=2\times (-g)\times h\]
Actually the velocity has two components v cos alpha and x Speed will be the magnitude
\[speed = \sqrt{(V \sin \alpha)^{2}-2gh +(V \cos \alpha)^2}\] \[= \sqrt{(V^2 (\sin ^2 \alpha +\cos^2 \alpha)-2gh}\] \[= \sqrt{V^2 -2gh}\]