Question

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Answer 1

$\bigstar\huge\red{\mathtt{Solution:}}$

(a)

Let the initial velocity of the projectile along the X and Y direction be V ox and V oy respectively

acceleration due to gravity 'g' is acting on the projectile in Y direction only

Let us consider  point C  as the highest point of the projectile

Let V bx and V by be the final velocity  of the projectile along the X and Y direction at point C

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Now in order to find the values of the final velocity at point C , let us use the first equation of motion ,

we know that ,

$\bigstar\boxed{\mathtt{V_b=V_o\underline{+}gt}}$

here,

• V b= final velocity
• V o= initial velocity
• g = acceleration due to gravity

• t = time taken

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Along X - direction :

$\rightarrow \mathtt{V_{bx}=V_{ox} - gt}$

we know that there is no accelerating force acting on the projectile in the X direction

$\rightarrow \mathtt{V_{bx}=V_{ox }- 0 \times t}$

$\rightarrow \mathtt{V_bx=V_ox } ....(1)$

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Along Y- direction :

$\rightarrow \mathtt{V_{by}=V_{oy} - gt} ......(2)$

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Now lets find the angle made by the projectile with the horizontal

we know that ,

$\rightarrow\mathtt{tan \theta =\dfrac{V _{by}}{V _{bx}}}$

From (1) and (2) we get ,

$\rightarrow\mathtt{tan \theta =\dfrac{V _{oy}- gt}{V _{ox}}}$

hence proved,

$\rightarrow\mathtt{\orange{\theta =tan^{-1} (\dfrac{V _{oy}- gt}{V _{ox}})}}$

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(b)

Let the maximum height of the projectile be( h m )

maximum height of the projectile is given by the formula ,

$\bigstar\boxed{\mathtt{h_m=\dfrac{u^{2} sin^{2} \theta}{2g}}}.....(1)$

Let the maximum range of the projectile be( R )

Range of the projectile is given by the formula ,

$\bigstar\boxed{\mathtt{R=\dfrac{u^{2} sin2 \theta}{g}}}$

we know that ,

$\bigstar\boxed{\mathtt{sin 2\theta =2 sin \theta .cos \theta }}$

$\bigstar\boxed{\mathtt{R=\dfrac{2u^{2} sin\theta.cos\theta}{g}}}.....(2)$

dividing (1) by (2) we get ,

$\rightarrow\mathtt{\dfrac{h_m}{R}=\dfrac{\dfrac{u^{2} sin^{2} \theta}{2g}}{\dfrac{2u^{2} sin\theta.cos\theta}{g}}}$

$\rightarrow\mathtt{\dfrac{h_m}{R}=\dfrac{u^{2} sin \theta.sin \theta}{2 g}\times\dfrac{g}{2u^{2} sin\theta.cos\theta}}$

$\rightarrow\mathtt{\dfrac{h_m}{R}=\dfrac{sin\theta}{4cos\theta}}$

$\rightarrow\mathtt{\dfrac{h_m}{R}=\dfrac{tan\theta}{4}}$

$\rightarrow\mathtt{tan\theta=\dfrac{4h_m}{R}}$

hence proved ,

$\rightarrow\mathtt{\blue{\theta=tan^{-1}(\dfrac{4h_m}{R})}}$

Answer 2

Answer:

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