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Hola ❤⏺ 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 ✒️ Show ur own talent ❌No spam ❌​

Answer»

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


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