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(B) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is t

Vedclass · Accepted Answer

$40$

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(B) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.For a fixed initial velocity $u$,the range $R$ is directly proportional to $\sin(2	heta)$.The range is maximum when $\sin(2	heta)$ is maximum,which occurs when $2	heta = 90^{\circ}$,or $	heta = 45^{\circ}$.We compare the values of $\sin(2	heta)$ for the given angles:For $25^{\circ}$,$2	heta = 50^{\circ}$,$\sin(50^{\circ}) \approx 0.766$.For $40^{\circ}$,$2	heta = 80^{\circ}$,$\sin(80^{\circ}) \approx 0.985$.For $55^{\circ}$,$2	heta = 110^{\circ}$,$\sin(110^{\circ}) = \sin(180^{\circ} - 110^{\circ}) = \sin(70^{\circ}) \approx 0.940$.For $70^{\circ}$,$2	heta = 140^{\circ}$,$\sin(140^{\circ}) = \sin(180^{\circ} - 140^{\circ}) = \sin(40^{\circ}) \approx 0.643$.Comparing these values,$\sin(80^{\circ})$ is the largest. Therefore,the range is maximum for the angle $40^{\circ}$.

Four projectiles are fired with the same velocities at angles $25^{\circ}, 40^{\circ}, 55^{\circ}$ and $70^{\circ}$ with the horizontal. The range of the projectile will be largest for the one projected at angle $.......$ . (in $^{\circ}$)

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