ᴅᴇꜰɪɴᴇ ᴩʀᴏᴊᴇᴄᴛɪʟᴇ ᴩᴀʀᴛɪᴄʟᴇ ᴀɴᴅ ᴅᴇʀɪᴠᴇ ɪᴛꜱ ᴇqᴜᴀᴛɪᴏɴ
Answers
Answer:
The projectile motion can be said to be the sum of two different types of motion, wherein the vertical component of the motion is an accelerated motion but the horizontal component of the motion is a motion with uniform velocity. Let us proceed with the derivation of the path of the projectile.
Formula Used: s=ut+1/2at^2 , displacement=velocity×time
Explanation:
The acceleration along the y-axis is given as −g where the negative sign denotes direction.
The horizontal component of the initial velocity of the body is vcosθ and the vertical component of the initial velocity is vsinθ
The displacement along the x-axis is given as X and the displacement along the y-axis is given as Y
Since the acceleration is acting along the y-axis, we’ll apply the second equation of motion to find the displacement.
The second equation of motion is given as s=ut+1/2at^2 where s is the displacement, u is the initial velocity and a is the acceleration
Substituting the values in the above equation, we get
Y=vsinθ(t)+1/2(−g)t^2⇒Y=vsinθt−gt^2/2−equation(1)
The motion along the x-axis is a motion with constant velocity so displacement can be given as displacement=velocity×time
Substituting the values, we get
X=vcosθ×t−equation(2)
Now we have to eliminate the time variable from the two equations. This is done by substituting the value of time from the second equation into the first equation. The equation can now be given as
Y=vsinθ(Xvcosθ)−1/2^g(Xvcosθ)^2⇒Y=Xtanθ−gX^2/2v^2cos^2θ
This is the required equation of projectile motion.
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PLS. MARK ME AS THE BRAINLIEST. HOPE THIS HELPS.
Answer:
hmm OK thanks ^_^
good evening anas ☺