A body is projected at an angle of 45^{circ} | vedclass

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Question

(D) Let the initial velocity be $u$ and the angle of projection be $	heta = 45^{\circ}$.The equation of the trajectory of a projectile is given by:$y

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) Let the initial velocity be $u$ and the angle of projection be $	heta = 45^{\circ}$.The equation of the trajectory of a projectile is given by:$y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$Given $x = 30 \,m$, $y = 20 \,m$, $	heta = 45^{\circ}$, and $g = 10 \,m/s^2$:$20 = 30 	an 45^{\circ} - \frac{10 	imes (30)^2}{2 u^2 \cos^2 45^{\circ}}$$20 = 30(1) - \frac{10 	imes 900}{2 u^2 	imes (1/2)}$$20 = 30 - \frac{9000}{u^2}$$\frac{9000}{u^2} = 10$$u^2 = 900 \implies u = 30 \,m/s$The horizontal range $R$ is given by:$R = \frac{u^2 \sin 2 	heta}{g} = \frac{900 	imes \sin 90^{\circ}}{10} = 90 \,m$The distance $s$ from the foot of the pole is the total range minus the initial distance from the pole:$s = R - 30 = 90 - 30 = 60 \,m$Thus, the correct option is $D$.

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