18. Advance Illustration | Kinematics | A Projectile Crossing Two Given Points | by Ashish Arora

18. Advance Illustration | Kinematics | A Projectile Crossing Two Given Points | by Ashish Arora


in this illustration we’ll analyze a projectile,
crossing 2 given points. we are given that a particle projected from origin at some angle
to horizontal passes through points ay b and b ay. referred to horizontal and vertical
axis through point of projections. we are required to find the horizontal range and
angle of projection in this situation. here, if we just analyze, there is a projectile
which is following a projectile trajectory and with initial speed u and, projected at
an angle theta. then in this situation if r, is the range of, this projectile motion.
and, we refer horizontal, as x axis vertical as y axis. in this equation we can write equation
of trajectory of. the particle is y is equal to x tan theta minus, g x square by 2 u square
coz square theta. here if we take this tan theta common out we have to multiplied by
sine theta, by sine theta and, the result can be re expressed this equation of trajectory,
can be re expressed as, x tan theta multiplied by 1 minus here 1 x and tan theta is taken
out. so this can be written as 1 minus, x by, range of projectile. as, we know the range
of, horizontal range of projectile is given as, u square, sine 2 theta by, g. so here
in denominator u square sine 2 theta by g i am writing as r. now in this situation if
there are 2 points. 1 is, ay comma b and say other is b comma ay. wherever these are located
randomly i have chosen the points. so both of these points must satisfy this trajectory
equation. so here we can write if projectile. passes. through ay comma b. and b comma ay.
we use here we can write b is equal to ay tan theta, multiplied by 1 minus, ay by r.
as well as we can write ay is equal to b tan theta multiplied by 1 minus, b by r. and,
in these 2 equations if we divide. here you can see if we, go for, this is first equation
this second equation we divided 1 by 2. see what we are getting. this will give us, b
square by, ay square is equal to here tan theta gets cancelled out. this r minus ay
upon, r minus b. if we further simplify this. it will give us b square, r minus b cube is
equal to ay square r minus. ay cube. or the range of, projectile we are getting here is
ay cube minus b cube divided by, a square minus, b square. which if we further simplify
will give us, ay square plus, ay b plus b square divided by, ay plus b. so 1 thing we
are required to find the horizontal range for which this is the answer. and we are also
required to find the angle of projection. so here, you can see as we got the value of
r. here, from equation 1. we can substitute the value of r over here see what we’ll
get. this b is, equal to ay. tan theta multiplied by 1 minus, ay by r so if we substitute the
value of r this is ay multiplied by ay plus b. divided by ay square plus ay b plus b square.
so if we simplify this expression the value of theta we are getting as tan inverse of,
ay square plus, ay b plus b square. divided by, ay b. this is another, result of this
problem.

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