Projectile Motion Video Lecture | Physics for EmSAT Achieve

projectile motion observe the athlete
throwing a javelin the javelin follows a
curved path and lands on the surface of
the earth such objects like javelin
which moves a longer path determined by
gravity and air resistance are called
projectiles the part followed by
projectile is called trajectory an
object dropped from a height an object
thrown vertically upwards and objects
thrown obliquely upwards are examples of
projectiles in all these cases for
influence of air resistance is neglected
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the velocity of the projectile varies
from point to point at each point the
velocity has components in the
horizontal and vertical directions the
horizontal component of velocity remains
unchanged throughout the flight the
vertical component of velocity
continuously changes due to the force of
gravity thus the horizontal motion is a
uniform motion and the vertical motion
is a uniformly accelerated motion the
magnitude of the vertical component is
different at different points and is
directed vertically downwards
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let us now analyze the motion of a prod
I'll thrown obliquely let a body be
projected from a point o with initial
velocity we owe the projectile makes an
angle theta with the horizontal axis the
velocity of the projectile at any point
of the trajectory goes on changing both
in magnitude and in direction the
velocity can be resolved into two
rectangular components we 0 cos theta
along x axis and we 0 sine theta along y
axis there is no acceleration in the
horizontal direction the vertical
component of acceleration ay y is equal
to minus g substituting the value of ax
and ay Y in equations 1 and 2
respectively
we get that is the vertical component of
velocity changes with time and at the
highest point the we Y becomes 0 let us
now find out the resultant velocity of
the projectile at any instant the
resultant velocity we is equal to vector
VX plus vector V y by parallelogram law
of vectors we is equal to square root of
VX squared plus V Y squared we is equal
to square root of V 0 cos theta whole
squared plus V 0 sine theta minus GT
whole squared the direction of the
project I'll at any instant is given by
the angle between the velocity vector
and the horizontal tan theta is equal to
we Y divided by we X from the equation V
Y is equal to V 0 sine theta minus GT it
is clear that we Y decreases when T
increases and becomes 0 at P at this
point the velocity is only along the
horizontal direction with further
increase in time the vertical component
of velocity gradually in
Risa's in the downward direction till
the particle strikes the ground
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horizontal range the distance from the
point
to the point where the projectile
strikes the horizontal plane is called
the horizontal range it is denoted by
the letter R
time of flight time of flight of a
projectile is the time taken by it from
the instant when it is projected to the
time when it strikes the point in a
horizontal plane passing through the
point of projection T is the equal to 2
v-0 sine theta by G
range of projectile range of projectile
is defined as the distance between the
point of projection and the point where
the trajectory cuts any plane r is equal
to we zero square sine 2 theta by G
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maximum height maximum height is the
maximum value of the vertical distance
attained by a projectile above the
horizontal plane to the point of
projection in the figure P is the point
of maximum height H is equal to V 0
square sine square theta by 2g
you