Question

help plzzzzzzzz...... ​

Answer 1

ɢɪᴠᴇɴ :-

The horizontal range is equal to the maximum height of a projectile motion .

ᴛᴏ ғɪɴᴅ :-

The horizontal range of the projectile.

ᴀɴsᴡᴇʀ :-

$\underline{\textsf{\textbf{\purple{\pink{ \leadsto }\:\: Diagram\:of\: projectile\:motion:-}}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0.2,0){\vector(0,1){5}}\put(3,0){\vector(1,0){3}}\put(3,0){\line(-1,0){2.8}}\put(0,-0.3){ \sf O }\qbezier(0.2,0)(2.5,4)(5.5,0)\put(5.5,-0.3){ \sf A }\put(2.75,0){\line(0,1){2}}\put(2.75 , 2){\vector( 1 ,0 ){1}}\put(0.7,0.6){\circle{0.4}}\put(2.74,2){\circle{0.4}}\put(5.1,0.6){\circle{0.4}}\put(2.75,-0.3){ \sf M }\put( 0.45,0.6){\vector(1,2){0.36}}\put( 5.3, 0.67){\vector(1, - 1){0.36}}\put( 2.2, - 0.45){\vector( - 1, 0){1.8}}\put( 3.3, - 0.45){\vector( 1, 0){2.2}}\put(2.3, - 0.5){ \tt Range }\put(2,2.5){ \tt Max^{m} height }\end{picture}$

Let us consider a ball projected at an angle θ , with the horizontal . Let us take the Range be R and Maximum height be h .

$\sf{\green{\hookrightarrow Firstly\:,\:in\:a\: projectile\:motion:-}}$

$\underline{\textsf{\textbf{\purple{\pink{ \leadsto }\:\: Maximum\:height\:is\:given\;by,}}}}$

$\boxed{\red{\bf \blue{\bigstar}\: H_{max}=\dfrac{u^2\sin^2\theta}{2g}}}$

$\rule{200}3$

$\underline{\textsf{\textbf{\purple{\pink{ \leadsto }\:\: Range\:is\:given\;by,}}}}$

$\boxed{\red{\bf \blue{\bigstar}\: Range=\dfrac{2u^2\sin2\theta}{g}}}$

$\rule{200}3$

$\underline{\pink{ \sf \longrightarrow Since\: they\:are\:equal :-}}$

$\tt:\implies H_{max}=Range$

$\tt:\implies \dfrac{u^2 sin^2\theta}{2g}=\dfrac{2u^2\sin 2\theta}{g}$

$\tt:\implies \dfrac{\cancel{u^2}\cancel{\sin^2\theta}}{2 \cancel g}=\dfrac{2\cancel{u^2}\cancel{\sin\theta}\cos\theta}{g}$

$\tt:\implies \dfrac{sin\theta}{2}=2\cos\theta$

$\tt:\implies \dfrac{sin\theta}{cos\theta}=2\times2$

${\boxed{\bf \red{\hookrightarrow} tan\theta = 4}}$

$\rule{200}3$

Now if we consider a right angled triangle ∆ABC in which tanθ = 4 . So using this we can find other trignometric values , here of cosθ and sinθ . And we know tanθ = $\sf \dfrac{perpendicular}{base}$ Now we need to find hypontenuse .

$\tt:\implies hypontenuse^2=base^2+perpendicular^2$

$\tt:\implies h^2 = 4^2+1^2$

$\tt:\implies h^2 = 16 + 1$

${\boxed{\bf \red{\hookrightarrow} hypontenuse= \sqrt{17}}}$

$\sf \orange{ So \: here , }$

$\sf\blue{ sin\theta=\dfrac{perp.}{hypo.}=\dfrac{4}{\sqrt{17}}}$

$\sf\blue{ cos\theta=\dfrac{base}{hypo.}=\dfrac{1}{\sqrt{17}}}$

Now , we will put these values in the formula of horizontal range , to get it .

$\rule{200}3$

$\underline{\textsf{\textbf{\purple{\pink{ \leadsto }\:\: Finding\:the\: horizontal\: range:-}}}}$

$\tt:\implies Horizontal\:range = \dfrac{2u^2\sin\theta\cos\theta}{g}$

$\tt:\implies R = \dfrac{2\times u^2 \times\dfrac{4}{\sqrt{17}}\times\dfrac{1}{\sqrt{17}}}{10}$

$\tt:\implies R = \dfrac{\cancel 2u^2\times 4}{17\times \cancel{10}}$

$\tt:\implies R = \dfrac{4u^2}{17\times 5}$

$\underline{\boxed{\red{\tt\longmapsto Range \:\:=\:\:\dfrac{4u^2}{85}}}}$

$\boxed{\green{\bf\pink{\dag}\:Hence\:the\: horizontal\:range\:is\:\dfrac{4u^2}{85}}}$

Answer 2

Answer:

ɢɪᴠᴇɴ :-

The horizontal range is equal to the maximum height of a projectile motion .

ᴛᴏ ғɪɴᴅ :-

The horizontal range of the projectile.

ᴀɴsᴡᴇʀ :-

Let us consider a ball projected at an angle θ , with the horizontal . Let us take the Range be R and Maximum height be h .

$\sf{\green{\hookrightarrow Firstly\:,\:in\:a\: projectile\:motion:-}}$

$\underline{\textsf{\textbf{\blue{\pink{ \leadsto }\:\: Maximum\:height\:is\:given\;by,}}}}$

$\boxed{\orange{\bf \pink{\bigstar}\: H_{max}=\dfrac{u^2\sin^2\theta}{2g}}}$

$\rule{200}3$

$\underline{\textsf{\textbf{\purple{\pink{ \leadsto }\:\: Range\:is\:given\;by,}}}}$

$\boxed{\orange{\bf \pink{\bigstar}\: Range=\dfrac{2u^2\sin2\theta}{g}}}$

$\rule{200}3$

$\underline{\blue{ \sf \longrightarrow Since\: they\:are\:equal :-}}$

$\tt:\implies H_{max}=Range$

$\tt:\implies \dfrac{u^2 sin^2\theta}{2g}=\dfrac{2u^2\sin 2\theta}{g}$

$\tt:\implies \dfrac{\cancel{u^2}\cancel{\sin^2\theta}}{2 \cancel g}=\dfrac{2\cancel{u^2}\cancel{\sin\theta}\cos\theta}{g}$

$\tt:\implies \dfrac{sin\theta}{2}=2\cos\theta$

$\tt:\implies \dfrac{sin\theta}{cos\theta}=2\times2$

${\boxed{\bf \red{\hookrightarrow} tan\theta = 4}}$

$\rule{200}3$

Now if we consider a right angled triangle ∆ABC in which tanθ = 4 . So using this we can find other trignometric values , here of cosθ and sinθ . And we know tanθ = $\sf \dfrac{perpendicular}{base}$ Now we need to find hypontenuse .

$\tt:\implies hypontenuse^2=base^2+perpendicular^2$

$\tt:\implies h^2 = 4^2+1^2$

$\tt:\implies h^2 = 16 + 1$

${\boxed{\bf \red{\hookrightarrow} hypontenuse= \sqrt{17}}}$

$\sf \orange{ So \: here , }$

$\sf\blue{ sin\theta=\dfrac{perp.}{hypo.}=\dfrac{4}{\sqrt{17}}}$

$\sf\blue{ cos\theta=\dfrac{base}{hypo.}=\dfrac{1}{\sqrt{17}}}$

Now , we will put these values in the formula of horizontal range , to get it .

$\rule{200}3$

$\underline{\textsf{\textbf{\purple{\pink{ \leadsto }\:\: Finding\:the\: horizontal\: range:-}}}}$

$\tt:\implies Horizontal\:range = \dfrac{2u^2\sin\theta\cos\theta}{g}$

$\tt:\implies R = \dfrac{2\times u^2 \times\dfrac{4}{\sqrt{17}}\times\dfrac{1}{\sqrt{17}}}{10}$

$\tt:\implies R = \dfrac{\cancel 2u^2\times 4}{17\times \cancel{10}}$

$\tt:\implies R = \dfrac{4u^2}{17\times 5}$

$\underline{\boxed{\pink{\tt\longmapsto Range \:\:=\:\:\dfrac{4u^2}{85}}}}$

$\boxed{\orange{\bf\pink{\dag}\:Hence\:the\: horizontal\:range\:is\:\dfrac{4u^2}{85}}}$