Question

Question

Determine the time of flight horizontal range and maximum height in projectile motion .

Answer ONLY Teacher , student of class 11/12

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Answer 1

Explanation:

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.

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Answer 2

$\bold{\underline{ \: You \: have \:to \: find }}$

1. $\bf{Time \: of \:Flight \: in \:projectile \: motion}$
2. $\bf{Horizontal \:range}$
3. $\bf{Maximum \:height}$

$\underline{{\boxed{ \sf\green{Brief \: Explaination \: and \:Solution} }}}$

Let a Projectile be projected with initial velocity (u) at angle ( θ) with horizontal direction.

The initial velocity acceleration in X and Y direction are :-

• $\bold{u_{x}}$ = u Cos θ , $\bold{u_{Y} }$= u Sin θ
• $\bold{a_{X}}$ = 0 , $\bold{a_{Y}}$ = g

$1. \underline{\bf{Time \: of \:Flight \: :-}}$

Let the time of flight of projectile is (T) , then it's displacement in Y direction in time T is 0 .

Then by second equation of Motion.

[S= ut+1/2 at²]

$\bold{ S= u_{Y} t + \frac{1}{2} a_{Y}t²}$

$\bold{ 0=( u Sinθ) T + \frac{1}{2} (- g)T²}$

$\bold{\frac{1}{2} gT \cancel{ {}^{2} } } = (\bold{u Sinθ}) • \cancel{ T}$

$\boxed{\bold{\green{T = \frac{2u Sin θ}{g}}}}$

$2. \underline{\bf{Horizontal \:range :- }}$

Let the Horizontal range of the projectile is (R) , but we know that

• $\boxed{ \sf{Displacement = Velocity × Time }}$

$\bold{Horizontal \: range = u_{x} × time\: of \: flight }$

$\bold{R = (u Cos θ) • (\frac{2 u Sin θ } {g} )}$

$\bold{R = \frac{u {}^{2} (2 sin θ • cos θ) }{g}}$ [ Sin 2θ = 2 Sin θ Cosθ]

$\boxed{ \green{\bold{R = \frac{u {}^{2} . sin 2θ }{g}}}}$

$3. \underline{\bf{Maximum \:height :- }}$

Let the maximum height of the projectile is H, but at maximum height the final Velocity in Y direction is 0.

by Third equation of motion [V² = u² + 2as]

$\bold{V{}^{2} Y = u{}^{2} + 2a_{Y}S}$

$\bold{0{}^{2} = ( u . Sinθ) {}^{2} + 2 (-g) • H}$

$\bold{2gH = {u}^{2} • {Sin}^{2} θ }$

$\boxed{\green{\bold{H = \frac{u² . Sin² θ }{2g}}}}$

$\pink{\bigstar} { \underline{\mathtt{\:Diagram}} }$

• See Above Attachment

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