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  • The ratio of a projectile's throw angle <img alt="(45^{\circ}-\theta)" src="https://entrancecorner.oncodecogs.com/gif.latex?%2845%5E%7B%5Ccirc%7D-%5Cthet

The ratio of a projectile's throw angle (45^{\circ}-\theta)  to an angle (45^{\circ}+\theta)  over the horizontal range described by the projectile is:

Option: 1

1:2


Option: 2

1:1


Option: 3

2:3


Option: 4

3:3


Answers (1)

best_answer

The projectile is projected at an angle (45^{\circ}-\theta) and   (45^{\circ}+\theta) , which are complementary to each other which means that the two angles add up to 90° . Hence, horizontal ranges will be equal.

R=u\ cos\alpha\times\frac{2u\ sin\alpha}{g}          

where \alpha is the angle of projection   

Or,  R=\frac{u^2sin2\alpha}{g}

For,    \alpha=45^{\circ}-\theta ,

we get:

        R_1=\frac{u^2sin(90^{\circ}-2\theta)}{g}

                = \frac{u^2\ cos2\theta}{g}  

    For \alpha=45^{\circ}+\theta

              R_2= \frac{u^2\ cos2\theta}{g}

Hence,           \frac{R_1}{R_2}=\frac{1}{1}

Or,     R_1:R_2=1:1

Posted by

Ritika Harsh

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