Question

Body of mass 2 Kg is projected at an angle of 45o with the horizontal. the maxim um height is 50 m then it horizontal range .
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Answer 1

Answer:

• $\textsf{Horizontal range of the body is 200m}$

Explanation:

${\underline{\underline{\sf{\red{{\bigstar}\:\:\:Given:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\green{ A \ body \ of \ mass \ 2kg \ is \ projected \ at \ an \ angle \ of \ 45^{o} \ with \ the \ horizontal}$

${\underline{\underline{\sf{\orange{{\bigstar}\:\:\:ToFind:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\purple{Horizontal \ range.}$

${\underline{\underline{\sf{\blue{{\bigstar}\:\:\: Solution:-}}}}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{Height_{\:(MAX)} = \dfrac{u^2\sin^2\theta}{2\times g} }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ 50=\dfrac{u^2 sin^2 45}{2 \times 9.8}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ 50=\dfrac{u^2 sin^245}{19.6}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{50\times 19.6 = u^2\times \dfrac{1}{\sqrt{2}} }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ 980 = \dfrac{u^2}{\sqrt{2}}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{u = 44.27 }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ Horizonal = \dfrac{(44.26)^2\times sin(45 \times 2)}{9.8}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ Horizonal = \dfrac{(44.26)^2\times sin(45 \times 2)}{9.8}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ Horizontal = \dfrac{1960 \times sin(45 \times 2)}{9.8}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\tt{ Horizontal = \dfrac{1960\times 1}{9.8}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:{\boxed{\tt{\red{ HORIZONTAL = 200 \; M}}}}$

Answer 2

Answer:

hmax=u²sin²thetha/2g

50=v²(1/2)(1/20)

u²=2000

range=u²sin2(thetha)/g

range=2000(1)/(10)

range=200m