Particle shot upward from ground??

Question

A particle is shot upward from the ground with a velocity of v and at an angle of θ above the horizontal.
Using vectos notations, what are the expressions for v and a just before it hits the ground again??

Answer 1

a remains unchanged throughout.
if Vi@theta is the initial velocity,
Vi@(-theta) will be the final velocity.

another way of stating it is vfy = -viy, vfx = vix

Answer 2

Nice one !!
we alwyas assume ideal conditions (no friction etc.)


i will try ot explain very simple describing the phenomenon.

the moment it leaves the ground with v (magnitude and angle => vector) it has a certain energy and a certain direction.
immidietly you ralise the two "components" vectors that make vector v
for the x axis is the Vx=cosθ * v and
for the y axis is the Vy=sinθ * v

for the x axis the vector remains the same for direction of course and for magnitude since doe x axis the body is moving with uniform speed. But for the Y axis the Vy changes direction (once) and changes its magnitude continously.
when it reaches the ground the Vx is the same andthe Vy has the same magnitude but opposite direction

Hmm.
so if i add up vectorially the Vx that doesnot cahenge at all and the original -Vy then we will ahve the complete v vector when the object returns to the groung

so

V final = v in magnitude

and θ new = 360-θ

and it all comes to it since they have the same cos and opposite sin

meaning that the Vy changes direction and the Vx remains the same.

Answer 3

a is a constant -9.8m/s^2. There is no acceleration in the horizontal direction. Assuming flat ground, v will have the same magnitude, but its direction will be 180-theta degrees.