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Derive an expression for maximum height and range of an object in projectile motion.

Answer
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Hint: As, here in this question, we need to derive the expression for maximum height and range of an object in projectile motion, we need to have a clear concept of the parabolic motion. We need to find out the trajectory or the path followed in a projectile motion. After that we need to use the components of the velocity vector in order to derive the expression for maximum height and range of an object in projectile motion. Also, we need to use Newton's equation of motion.

Complete answer:
Step one
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The horizontal component of velocity is given by, vx=ucosθ
The vertical component of the velocity is given by, vy=usinθ
Let us assume that the body reaches the point P(x,y), after time t
Also, let us assume the maximum height to be H
Now, if we need to find the Horizontal distance, we can write the equation for it as,
x=ucosθt ………(i) As, distance = speed ×time
Step two
Now, for maximum height of an object for projectile motion can be found by using third equation of motion, v2u2=2as
So, putting the values in the above equation, we get,
o2(usinθ)2=2(g)H
H=u2sin2θ2g
H=u2sin2θ2g
Step three
Let the horizontal range be R
We know that time of flight, T=2usinθg
Now, using equation (i) we can write,
R=ucosθT
R=ucosθ×2usinθg
R=u2sin2θg
Hence, the required value of maximum height is u2sin2θ2g and the range is u2sin2θg.

Note:
We should not confuse time of maximum height with time of flight. Time of maximum height is the time when the object attains the maximum height and is given by t=usinθg. Time of flight is the total time taken by the object to cover the total horizontal distance or in other words the time till when the object is in air and is given by T=2usinθg.