Question

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​

Answer 1

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

Given :-

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

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To Find :-

Maximum height

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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

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

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

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

Answer 2

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

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

• 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.

$\rule{300}{1.5}$