A projectile is fired at an angle of 30^{circ | vedclass

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(D) Given: Angle of projection $	heta = 30^{\circ}$,vertical component of initial velocity $u_y = 80\,m/s$.$1$. Find horizontal component $u_x$:$	an

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(D) Given: Angle of projection $	heta = 30^{\circ}$,vertical component of initial velocity $u_y = 80\,m/s$.$1$. Find horizontal component $u_x$:$	an 	heta = \frac{u_y}{u_x} \implies 	an 30^{\circ} = \frac{80}{u_x} \implies \frac{1}{\sqrt{3}} = \frac{80}{u_x} \implies u_x = 80\sqrt{3}\,m/s$.$2$. Find time of flight $T$:$T = \frac{2u_y}{g} = \frac{2 	imes 80}{10} = 16\,s$.$3$. Find velocity at $t = \frac{T}{4} = \frac{16}{4} = 4\,s$:Horizontal velocity $v_x = u_x = 80\sqrt{3}\,m/s$.Vertical velocity $v_y = u_y - gt = 80 - (10 	imes 4) = 80 - 40 = 40\,m/s$.$4$. Magnitude of velocity $v$:$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(80\sqrt{3})^2 + (40)^2} = \sqrt{6400 	imes 3 + 1600} = \sqrt{19200 + 1600} = \sqrt{20800} = 40\sqrt{13} \approx 144.22\,m/s$.Since $144.22\,m/s$ is not among the options,the correct choice is $(D)$.

$A$ projectile is fired at an angle of $30^{\circ}$ to the horizontal such that the vertical component of its initial velocity is $80\,m/s$. Its time of flight is $T$. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly $........\,m/s$.

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