Question

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

Answer 1

@UnkleRhaukus @ash2326

Answer 2

@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}

Do you understand this part?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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