Question

Show that the range is same the two angle of projection that is 30 and 60 degrees with the same initial velocity of 10m/s . when g= 10 m/s.

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Answer 1

Answer:

Given :

Initial velocity ( u ) = 10 m / s

We have to take value of g = 10 m / sec²

We know :

Range ( R ) = u² sin 2 θ / g

First  [ when θ = 30 ]

R = 10² × sin 2 × 30 / 10

R = 10 × sin 60     [ sin 60 = √ 3 / 2 ]

R = 5 √ 3 m

Second [ when θ = 60 ]

R' = 10² × sin 2 × 60 / 10

R' = 10 × sin 120    [ sin ( 90 + 30 ) = cos 30 ]

R' = 10 × √ 3 / 2

R' = 5 √ 3 m

Now clearly:

R = R'

Hence shown .

Answer 2

Answer:-

Yes, The range is same at 30° and 60°

Explanation:

Given :-

initial velocity (v)= 10 m/s

Formula used:-

$\bf{Range \: = \frac{ {v}^{2}sin \: 2 \theta }{g} }$

Solution:-

(1) when angle of projection is 30°

Then range is→

$\bf{R1 = \frac{ {(10)}^{2} sin \: 2 \times 30 \degree}{g} } \\ \\ \bf{assuming \: g = 10 \: \frac{m}{s} } \\ \\ \bf{R1 = \frac{100 \times sin \: 60 \degree}{10} } \\ \\ \bf{R1 = 10 \times \frac{ \sqrt{3} }{2} = 5 \sqrt{3} }$

Now ,

(2) when angle of projection is 60°

Then range is →

$\bf{R2 = \frac{ {(10)}^{2} sin \: 2 \times 60 \degree}{10} } \\ \\ \bf{R2 = \frac{100 \times sin \: 120 \degree}{10} } \\ \\ \bf{R2 = 10 \: sin \: 120 \degree} \\ \\ \bf{ R2 = 10 \times \frac{ \sqrt{3} }{2} } \: = 5 \sqrt{3}$

Therefore ,R1 = R2

Hence, the range is same the two angle of projection that is 30° and 60 °