Physics, asked by nishu884, 11 months ago

a projectile is fired at such an angle as to have a maximum horizontal range of 5000 m find the maximum height attained by it. g=9.8 m/ s^2​

Answers

Answered by Anonymous
13

\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}

Given :-

  • Maximum Horizontal range is 5000 m
  • g = 9.8 m/s²

===================================

To Find :-

Maximum height

===================================

Solution :-

We have formula for range :-

\LARGE {\boxed{\boxed{\sf{R \: = \: \frac{u^2  \: Sin \: 2 \theta }{g}}}}}

As we are given maximum range is 5000 m and for maximum range θ = 45°

So, putting values

\Large \leadsto {\sf{5000 \: = \: \frac{u^2 \: Sin \: 2(45^{\circ})}{9.8}}}

\Large \leadsto {\sf{u^2 \: = \: \frac{5000 \:  \times \: 9.8}{Sin \: 90^{\circ}}}}

As value of Sin90° is 1

\Large \leadsto {\sf{u^2 \: = \: \frac{4900}{1}}}

\Large \leadsto {\sf{u \: = \: \sqrt{49000}}}

\huge \implies {\boxed{\boxed{\green{\sf{u \: = \: 221.35 \: ms^{-1}}}}}}

\rule{200}{2}

Now, formula for maximum height is

\LARGE {\boxed{\boxed{\sf{h \: = \: \frac{u^2 \: Sin^2 \theta }{2g}}}}}

Same here θ will be 45°

\Large \leadsto{\sf{h \: = \: \frac{(221.35)^2 \: \times \:(\frac{1}{\sqrt{2}})^2}{2 \: \times \: 9.8}}}

As value of Sin45° is 1/√2

\Large \leadsto{\sf{h \: = \: \frac{{\cancel{49000}} \: \times \: \frac{1}{\cancel{2}}}{{19.6}}}}

\Large \leadsto {\sf{h \: = \: \frac{\cancel{24500}}{\cancel{19.6}}}}

\huge \implies {\boxed{\boxed{\green{\sf{h \: = \: 1250 \: m}}}}}


αmαn4чσu: Awesome answer :)
Anonymous: Great answer
Answered by ShivamKashyap08
27

\huge{\bold{\underline{\underline{....Answer....}}}}

\huge{\bold{\underline{Given:-}}}

  • Horizontal Range (R) = 5000 meters.
  • Acceleration due to gravity (g) = 9.8 m/s².

\huge{\bold{\underline{Explanation:-}}}

\rule{300}{1.5}

From the Range formula ,

\large{\boxed{\tt R = \dfrac{u^2sin2\theta}{g}}}

Substituting the values,

\large{\tt \leadsto 5000 = \dfrac{u^2 sin2 \theta}{9.8}}

For Range to be maximum

∴sin2θ = 1. [Here θ = 45°]

Substituting it,

\large{\tt \leadsto 5000 = \dfrac{u^2 \times 1}{9.8}}

\large{\tt \leadsto 5000 = \dfrac{u^2 }{9.8}}

\large{\tt \leadsto 5000 \times 9.8 = u^2}

\large{\tt \leadsto u^2 = 49000}

\large{\tt \leadsto u = \sqrt{49000}}

\large{\boxed{\tt u = 221.35 \: m/s}}

So, the initial velocity is 221.35 m/s.

\rule{300}{1.5}

\rule{300}{1.5}

Now, Applying Maximum Height Formula,

\large{\boxed{\tt H_{max} = \dfrac{u^2sin^2\theta}{2g}}}

Substituting the values,

\large{\tt \leadsto  H_{max} = \dfrac{(221.35)^2 \times sin^2 45 \degree}{2 \times 9.8}}

\large{\tt \leadsto  H_{max} = \dfrac{49000 \times \left( \dfrac{1}{\sqrt{2}} \right)^2}{19.6}}

\large{\tt \leadsto  H_{max} = \dfrac{49000 \times \dfrac{1}{2}}{19.6}}

\large{\tt \leadsto H_{max} = \dfrac{\cancel{49000} \times \dfrac{1}{\cancel{2}}}{19.6}}

\large{\tt \leadsto  H_{max} = \dfrac{24500}{19.6}}

\large{\tt \leadsto H_{max} = \dfrac{\cancel{24500}}{\cancel{19.6}}}

\huge{\boxed{\boxed{\tt H_{max} = 1250 \: meters}}}

So, the Maximum height reached is 1250 meters.

\rule{300}{1.5}


αmαn4чσu: Nice answer as always ✌
Anonymous: Great answer
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