Kinematics 2D - Projectile from a Height at an Angle Θ

?Kinematics 2D - Motion taking place in 2 dimensions (x and y)

This is the concept -?Projectile from a height at an angle theta.

This concept stumped me as I could not find a derivation for its formula and to be honest I could not find a formula to find it's?"Range"?and?"Time Taken". So I stood up to the challenge and tried to a solution to this problem of mine.

I know this type of projectile problems can be done with the "Equation of Trajectory", but I hope this too helps you to solve those types of questions.

First of all you should that the ball is projected with an angle theta (the angle between the direction of projection and the x axis), so we can break it up into its two components in the x and y axes:

1. u.sin theta?along the y-axis?????????????????????????????????2. u.cos theta?along the x axis

Now the thing I did was this:

So what I did was that, I took the highest point of the entire projectile and cut the projectile through that point. Hence the projectile from a height at an angle gets divided into two projectile at a height.

Projectile A - Projectile from a height

height - H(α), velocity - u.cos theta and???acceleration - gravity (g)

h + H = H(α)

For Projectile A:

1. Time of Flight for A :-?T(A) = √[2.H(α)/g]

2. Range for A :-?R(A) = ucosΘ.√[2.H(α)/g]

Well we got this for projectile A

Go down to see Projectile B

Projectile B is the reverse of projectile of the ball from the highest point to point 'a'.

height - h, velocity - u.cos theta and??????????acceleration - gravity (g)

Therefore for Projectile B:

1. Time of Flight for B :-?T(B) = √[2.h/g]

2. Range for B :-?R(B) = ucosΘ.√[2.h/g]

Now that we have found formulae for both the projectiles, let's combine them.

Projectile A + Projectile B = Final/Original Projectile

T(A) + T(B) = Final Time of Flight

T(A) = √[2.H(α)/g] and T(B) = √[2.h/g]

√[2.H(α)/g] + √[2.h/g] = √(2/g) [ {√H(α)} + {√h} ]

but wait.... wait wait wait.

How do we find 'h'.

The max height attained during a projectile is only due to the y-component of the initial velocity and using the formula

v^2 = u^2 + 2as;??final velocity = v = 0m/s at the max height (VERTICAL MOTION) and?acceleration (gravity)?and?displacement (height)?are in different directions

Therefore, 0 = u^2 - 2gh;???u^2 = 2gh;?????height = u^2 / 2g

therefore - the maximum height in the projectile will be

h (to the highest point) = [u^2.sin(^2)Θ / 2g]

by the way, H will be normally given in the question, and now that you know how to find 'h', we can find H(α) by adding the two.

Now knowing every single quantity we needed for the entire Projectile, now we have:

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Time of Flight (T) = √(2/g) [ {√H(α)} + {√h} ]

????????????????????? =??√[2.H(α)/g] + u.sin theta/g

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Range (R) = R(A) + R(B)

Range (R) = ucosΘ.√[2.H(α)/g] + ucosΘ.√[2.h/g]

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Range (R) = ucosΘ.√{2/g}?[ {√H(α)} + {√h} ]

???????????? ?=?ucosΘ.√[2.H(α)/g] + [u(^2).cosΘsinΘ]/g

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Final Formulae

• h (to the highest point) = [u^2.sin(^2)Θ / 2g]
• Time of Flight (T) = √(2 / g) [ {√ H(α)} + {√ h} ]
• Range (R) = u.cos theta.√{2 / g}?[ {√ H(α)} + {√ h} ]
• Range (R) = u.cos theta x Time of Flight (T)

Note from the writer:

I don't know if there is a simpler approach to such questions or if this formula is already out there but this is the first time I found it, my only hope is that this is an addition to your understanding of 2 dimensional motion.