Question

the que. no. 4 is from projectile motion yes or no

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Answer 1

Answer:

1 : 1

Option 1. is correct.

Explanation:

We have formula for Range ( R )

$\large R=\dfrac{u^2\sin2\theta}{g}$

where u = initial velocity

g = acceleration due to gravity

θ = angle

Now take ratio for both angle i.e. ( 45 - θ  )  and ( 45 + θ )

$\large \text{ \dfrac{R}{R^'} =\left(\dfrac{\dfrac{u^2\sin2\theta}{g}}{\dfrac{u^2\sin2\theta}{g}} \right) }\\\\\\\large \text{ \dfrac{R}{R^'} =\left(\dfrac{\dfrac{\cancel u^2\sin2(45-\theta)}{\cancel g}}{\dfrac{\cancel u^2\sin2(45+\theta)}{\cancel g}} \right) }\\\\\\\large \text{ \dfrac{R}{R^'} =\left(\dfrac{\dfrac{\sin(90-2\theta)}{1}}{\dfrac{\sin(90+2\theta)}{1}} \right) }\\\\\\\large \dfrac{R}{R^'} =\left(\dfrac{\sin(90-2\theta}{\sin(90+2\theta}\right)$

Let 2 θ = α

$\large \dfrac{R}{R^'} =\left(\dfrac{\sin(90-\alpha}{\sin(90+ \alpha}\right)$

We know that  sin ( 90 -  θ ) = cos  θ and sin ( 90 +  θ ) = cos  θ

$\large \dfrac{R}{R^'} =\left(\dfrac{\cos\alpha}{\\cos\alpha}\right)\\\\\\\large \dfrac{R}{R^'} =\left(\dfrac{1}{1}\right)$

Thus we get ratio 1 : 1.