Question

for an object thrown 45° to horizontal the maximum height H and horizontal range R are releated as​

Answer 1

Answer:

Given:

Object thrown (projected) at 45°

To find:

Relationship between H and R

Calculation:

For 45° angle of Projection, we get max value of Range (R) ;

$R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

Putting available values :

$= > R = \dfrac{ {u}^{2} \sin(2 \times 45 \degree) }{g}$

$= > R = \dfrac{ {u}^{2} \sin( 90 \degree) }{g}$

$= > R = \dfrac{ {u}^{2} }{g} .....(1)$

Coming to max Height :

$H = \dfrac{ {u}^{2} { \sin}^{2} (\theta) }{2g}$

Putting available values:

$= > H = \dfrac{ {u}^{2} { \sin}^{2} (45 \degree) }{2g}$

$= > H = \dfrac{ {u}^{2} { (\frac{1}{ \sqrt{2} } )}^{2} }{2g}$

$= > H = \dfrac{1}{4} \times \bigg(\dfrac{ {u}^{2} }{g} \bigg)$

$= > H = \dfrac{R}{4}$

So relationship is as follows:

$\boxed{ \red{ \huge{ \bold{H = \dfrac{R}{4} }}}}$

Answer 2

Solution :

⏭ Given:

✏ Angle of projection = 45°

⏭ To Find:

✏ Relation between maximum height (H) and horizontal range (R)

⏭ Formula:

• Maximum height

$\star \: \boxed{ \bold{ \tt{ \pink{H = \dfrac{ {u}^{2} { \sin}^{2} \theta }{2g}}}}} \: \star$

• Horizontal range

$\star \: \boxed{ \bold{ \green{ \tt{R = \dfrac{ {u}^{2} \sin2 \theta}{g}}}}} \: \star$

⏭ Calculation:

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• Maximum height

$\dashrightarrow \sf \: H = \dfrac{ {u}^{2} { \sin}^{2} 45 \degree }{2g} \\ \\ \dashrightarrow \sf \: \red{H = \dfrac{ {u}^{2} }{4g} }$

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• Horizontal range

$\rightarrowtail \sf \: R = \dfrac{ {u}^{2} \sin2(45 \degree)}{g} \\ \\ \rightarrowtail \sf \: R = \dfrac{ {u}^{2} \sin90 \degree }{g} \\ \\ \rightarrowtail \sf \: \blue{R = \dfrac{ {u}^{2} }{g} }$

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$\mapsto \sf \: \dfrac{R}{H} = \dfrac{ {u}^{2}}{g} \times \dfrac{4g}{ {u}^{2} } \\ \\ \mapsto \sf \: \frac{R}{H} = 4 \\ \\ \mapsto \: \boxed{ \tt{ \huge{\orange{R = 4H}}}}$