Question

. A body is projected at angle 30° to the
horizontal with a velocity 50 ms-1 maximum
height of projectle is​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\to \textsf{ A body is projected at an angle of \sf 30^{\circ} .}\\\to\textsf{ It is projected at an angle of 50m/s .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\to \textsf{ The maximum height of the projectile .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

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

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

The ball is projected at an angle of 30° and the ball is projected with a velocity of 50m/s. So for a projectile motion the maximum height is given by ,

$\sf:\implies\pink{ Height_{max}= \dfrac{u^2\ sin^2\theta}{2g}} \\\\\sf:\implies H_{max} = \dfrac{ (30 m/s)^2 sin^2 (30^{\circ}) }{2\times 10m/s^2}\\\\\sf:\implies H_{max}= \dfrac{ 900m^2/s^2 \times \bigg(\dfrac{1}{2}\bigg)^2 }{ 20m/s^2}\\\\\sf:\implies H_{max}= \dfrac{ 900 \times \dfrac{1}{4}}{20} m \\\\\sf:\implies H_{max}= \dfrac{ 900 }{4\times 20} \\\\\sf:\implies\boxed{\pink{\mathfrak{ Height_{(maximum)} = 11.25 \ m }}}$

$\underline{\underline{\textsf{\textbf{ \blue{ \therefore Hence the maximum height is 11.25 m . }}}}}$