Question

Abody is projected with velocity u such that its horizontal range and maximum vertical heights are same.
Then the maximun heights is​

Answer 1

Answer:

Given:

A body is projected such that range and max vertical height is same.

To find:

Max height

Formulas used:

Range = u²sin(2θ)/g

Max height = u²sin²(θ)/2g

Calculation:

As per question,

Range = Max height

=> u²sin(2θ)/g = u²sin²(θ)/2g

=> sin(2θ) = sin²(θ)/2

=> 2 sin(θ) cos(θ) = sin²(θ)/2

=> tan(θ) = 4

So sin(θ) = 4/√17

Now , max height = u²sin²(θ)/2g

=> Max height = u² (4/√17)²/2g

=> Max height = u²/g × 16/17 × 1/2

=> Max height = 8u²/17g.

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A body is projected with velocity u such that its horizontal range and maximum vertical heights are same. Then the maximun heights is?

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

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

• Let the angle at the Projectile is projected be θ.
• The range & Height of Projectile be "R" and "H" respectively. 

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

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As Mentioned in the above Question,

$\large{\boxed{\tt H_{max} = R}}$

Substituting the Respective Formulas,

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2g} = \dfrac{u^2 sin 2\theta}{g}}$

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2\cancel{g}} = \dfrac{u^2 sin 2\theta}{\cancel{g}}}$

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2} = u^2 sin 2\theta}$

As we Know,Sin2θ = 2sinθcosθ,

Substituting it,

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2} = u^2 \times 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{\cancel{u^2} sin^2 \theta}{2} = \cancel{u^2} \times 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{sin^2 \theta}{2} = 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{sin\theta \times sin\theta}{2} = 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{sin\theta \times \cancel{sin\theta}}{2} = 2\cancel{sin \theta} cos \theta}$

$\large{\tt \leadsto \dfrac{sin\theta }{2} = 2 cos \theta}$

Rearranging,

$\large{\tt \leadsto \dfrac{sin \theta}{cos \theta} = 2 \times 2}$

As we know,Sinθ/Cosθ = Tanθ,

Substituting it, we get

$\large{ \leadsto {\underline{\underline {\tt Tan\theta = 4}}}}$

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Now, Using Trigonometric identities,

$\large{\boxed{\tt Sec^2 \theta = 1 + tan^2 \theta}}$

Substituting tanθ value,

$\large{\tt \leadsto Sec^2 \theta = 1 + (4)^2}$

$\large{\tt \leadsto Sec^2 \theta = 1 + 16}$

$\large{\tt \leadsto Sec^2 \theta = 17}$

As we know,

$\large{\tt Cos^2 \theta = \dfrac{1}{Sec^2 \theta}}$

Now,

$\large{\tt \leadsto Cos^2 \theta = \dfrac{1}{17}}$

Applying First Trigonometric Identity.

$\large{\boxed{\tt Sin^2 \theta + Cos^2 \theta = 1}}$

$\large{\tt \leadsto Sin^2 \theta + \dfrac{1}{17} = 1 }$

$\large{\tt \leadsto Sin^2 \theta = 1 - \dfrac{1}{17}}$

$\large{\tt \leadsto Sin^2 \theta = \dfrac{17 - 1}{17}}$

$\large{ \leadsto {\underline{\underline {\tt Sin^2\theta = \dfrac{16}{17}}}}}$

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Now, Applying Maximum Height,

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

$\large{\tt \leadsto H_{max} = \dfrac{u^2}{2g} \times Sin^2 \theta}$

Substituting the values,

$\large{\tt \leadsto H_{max} = \dfrac{u^2}{2g} \times \dfrac{16}{17}}$

$\large{\tt \leadsto H_{max} = \dfrac{u^2}{\cancel{2}g} \times \dfrac{\cancel{16}}{17}}$

Simplifying gives,

$\huge{\boxed{\boxed{\tt H_{max} = \dfrac{8u^2}{17g} }}}$

Hence derived!

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