Question

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⏺ A projectile is projected from a point 5m away from the foot of 10 m high tower such that it touches the tower and falls on other sides of it at a distance 15m from foot of tower . calculate :-

a) R b) H c) T d) U v) Theta

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Answer 1

Answer:

1. R = 20 m

2. H = 10 m

3. T = 4√5/√g sec.

4. u = 5√g m/s

5. $\theta={\tan}^{-1}2$

Explanation:

Let the projectile is projected at an angle $\alpha$ with the horizontal with an initial velocity u.

Now, it's given that,

It's projected from a point at a distance of 5 m from the foot of a tower which is 10 m high.

Also, it just touches the tower and falls to a distance of 15 m from the foot of the tower.

Therefore, total horizontal distance covered by the projectile = 5 + 15 = 20 m

But, we know that, maximum horizontal distance is called range.

Therefore, we will get,

(1) Range, R = 20 m

Also, it just touches the highest point of tower and falls after that.

Therefore, maximum height attained is,

(2) Maxm. height, H = 10 m

Now, we know that,

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

Therefore, we will get,

$= > \dfrac{2 {u}^{2} \sin( \alpha ) \cos( \alpha ) }{g} = 20 \\ \\ = > \dfrac{2 {u}^{2} { \sin}^{2} \alpha \cos\alpha }{2g \sin( \alpha ) } = \dfrac{20}{2} \\ \\ = > \frac{ {u}^{2} { \sin }^{2} \alpha }{2g} \times \cot( \alpha ) = 5$

But, we know that,

$\dfrac{ {u}^{2} { \sin }^{2} \alpha }{2g} = H$

And, H = 10 m

Therefore, substituting the values, we get,

$= > \cot( \alpha ) = \frac{5}{10} \\ \\ = > \cot( \alpha ) = \frac{1}{2} \\ \\ = > \tan( \alpha ) = 2 \\ \\ = > \alpha = { \tan}^{ - 1} 2$

Therefore, angle at which it is projected,

(5) $\bold{\theta={\tan}^{-1}2}$

Now,

we know that,

$\sin(2 \alpha ) = \dfrac{2 \tan( \alpha ) }{1 + { \tan }^{2} \alpha }$

Substituting the values, we get,

$= > \sin(2 \alpha ) = \dfrac{2 \times 2}{1 + {2}^{2} } \\ \\ = > \sin(2 \alpha ) = \dfrac{4}{1 + 4} \\ \\ = > \sin(2 \alpha ) = \dfrac{4}{5}$

Now, we have,

$= > \dfrac{ {u}^{2} \sin(2 \alpha ) }{g} = 20 \\ \\ = > {u}^{2} \times \dfrac{4}{5} = 20g \\ \\ = > {u}^{2} = \frac{5 \times 20}{4}g \\ \\ = > {u}^{2} = 25g \\ \\ = > u = 5 \sqrt{g}$

Therefore, the projectile is projected with velocity equal to ,

(4) u = 5√g m/s.

Now, we know that,

Time of flight, $t=\dfrac{2u\sin\alpha}{g}$

Therefore, substituting the values, we get,

=> t = 2u/g × 2/√5

=> t = 4u/√5g

=> t = 4× 5√g/√5g

=> t = 4√5/√g

Hence, we get,

(3) T = 4√5/√g

Answer 2

Solution :

⏭ Given:

✏ A projectile is projected from a point 5m away from the foot of 10m high tower such that it touches the tower and falls on other sides of it at a distance 15m from foot of tower.

✏ Please see the attachment for better understanding.

⏭ To Find:

• Range
• Height
• Total time of flight
• Initial velocity
• Angle of projection

⏭ Formula:

$\bigstar \tt \: \red{Range(R) = \dfrac{ 2{u}^{2} \sin \theta \cos \Theta}{g}} \\ \\ \bigstar \tt \: \blue{Height(H) = \dfrac{ {u}^{2} { \sin}^{2} \Theta}{2g}} \\ \\ \bigstar \tt \: \green{Time \: of \: flight(T) = \dfrac{2u \sin \Theta}{g} }$

✏ Relation between range and height is given by

$\bigstar \boxed{ \tt{\pink{R \tan \Theta = 4H}}}$

⏭ Calculation:

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1) Max. Range

$\implies \sf \: Range = AO + OB \\ \\ \implies \sf \: range = 5 + 15 \\ \\ \implies \: \boxed{ \tt\red{Range = 20 \: m}}$

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2) Max. height

$\mapsto \sf \: Max. \: Height = OM \\ \\ \mapsto \: \boxed{ \tt\blue{Max. \: Height = 10 \: m}}$

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3) Angle of projection

$\leadsto \sf \: R \tan \Theta = 4H \\ \\ \leadsto \sf \: \tan \Theta = \dfrac{4H}{R} \\ \\ \leadsto \sf \: \tan \Theta = \dfrac{4 \times 10}{20} \\ \\ \leadsto \sf \: \tan \Theta = 2 \\ \\ \leadsto \: \boxed{ \tt\green{ \Theta = { \tan}^{ - 1} (2)}}$

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4) Initial velocity

$\dashrightarrow \sf \: R = \dfrac{ 2{u}^{2} \sin \Theta \cos \Theta }{g} \\ \\ \dashrightarrow \sf \: 20= \dfrac{2 {u}^{2} \times \dfrac{2}{ \sqrt{5} } \times \dfrac{1}{ \sqrt{5} } }{g} \\ \\ \dashrightarrow \sf \: 20g = {u}^{2} \times \dfrac{4}{5} \\ \\ \dashrightarrow \sf \: {u}^{2} = 25g \\ \\ \dashrightarrow \: \boxed{ \tt\purple{u = 5 \sqrt{g} \: mps }}$

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5) Time of flight

$\rightarrowtail \sf \: T = \dfrac{2u \sin \Theta}{g} \\ \\ \rightarrowtail \sf \: T = \dfrac{2 \times 5 \sqrt{g} \times \dfrac{2}{ \sqrt{5} } }{g} \\ \\ \rightarrowtail \sf \: T = \dfrac{2 \times 2 \times \sqrt{5} \times \cancel{\sqrt{5} \times \sqrt{g}} }{ \cancel{\sqrt{5} \times \sqrt{g} } \times \sqrt{g} } \\ \\ \rightarrowtail \: \boxed{ \tt\orange{Time \: of \: flight =4 \sqrt{ \dfrac{5}{g} } \: s}}$

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