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(B) Let the initial velocity be $u$ and the angle of projection be $	heta$.The horizontal component of velocity is $v_x = u \cos 	heta$ (constant).T

Vedclass · Accepted Answer

$	an^{-1}(2)$

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(B) Let the initial velocity be $u$ and the angle of projection be $	heta$.The horizontal component of velocity is $v_x = u \cos 	heta$ (constant).The vertical component of velocity at time $t$ is $v_y = u \sin 	heta - gt$.Given that at $t = 1 \, s$,the angle with the horizontal is $45^{\circ}$:$	an 45^{\circ} = \frac{v_y}{v_x} = \frac{u \sin 	heta - g(1)}{u \cos 	heta} = 1$$u \sin 	heta - g = u \cos 	heta \implies u \sin 	heta - u \cos 	heta = g \quad \dots (1)$Given that the speed is minimum at $t = 2 \, s$. The speed of a projectile is minimum at the highest point where the vertical component of velocity is zero $(v_y = 0)$:$u \sin 	heta - g(2) = 0 \implies u \sin 	heta = 2g \quad \dots (2)$Substitute $(2)$ into $(1)$:$2g - u \cos 	heta = g \implies u \cos 	heta = g \quad \dots (3)$Dividing $(2)$ by $(3)$:$\frac{u \sin 	heta}{u \cos 	heta} = \frac{2g}{g} \implies 	an 	heta = 2$$	heta = 	an^{-1}(2)$.

$A$ particle projected from the ground moves at an angle of $45^{\circ}$ with the horizontal one second after projection,and its speed is minimum two seconds after the projection. What is the angle of projection of the particle? [Neglect the effect of air resistance]

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