A body is projected at an angle of 60^{circ} | vedclass

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Question

(C) Let the initial velocity be $v$ and the angle of projection be $	heta = 60^{\circ}$.Given,the vertical component of initial velocity $v_y = v \si

Vedclass · Accepted Answer

$30.54$

Vedclass · Answer

(C) Let the initial velocity be $v$ and the angle of projection be $	heta = 60^{\circ}$.Given,the vertical component of initial velocity $v_y = v \sin 	heta = 40 \ m \ s^{-1}$.The time of flight $T = \frac{2 v \sin 	heta}{g} = \frac{2 	imes 40}{10} = 8 \ s$.The time at which we need to find the velocity is $t = \frac{T}{4} = \frac{8}{4} = 2 \ s$.The horizontal component of velocity remains constant throughout the motion: $v_x = v \cos 	heta$.Since $v \sin 	heta = 40$,we have $v = \frac{40}{\sin 60^{\circ}} = \frac{40}{\sqrt{3}/2} = \frac{80}{\sqrt{3}} \approx 46.19 \ m \ s^{-1}$.Thus,$v_x = \frac{80}{\sqrt{3}} 	imes \cos 60^{\circ} = \frac{80}{\sqrt{3}} 	imes \frac{1}{2} = \frac{40}{\sqrt{3}} \approx 23.09 \ m \ s^{-1}$.The vertical component of velocity at time $t$ is $v_y(t) = v \sin 	heta - g t = 40 - (10 	imes 2) = 20 \ m \ s^{-1}$.The magnitude of velocity at time $t$ is $v_t = \sqrt{v_x^2 + v_y(t)^2} = \sqrt{(23.09)^2 + (20)^2} = \sqrt{533.15 + 400} = \sqrt{933.15} \approx 30.54 \ m \ s^{-1}$.

Column $I$	Column $II$
$(A)$ Horizontal component of velocity	$(p)$ $5$ $SI$ unit
$(B)$ Vertical component of velocity	$(q)$ $10$ $SI$ unit
$(C)$ Horizontal displacement	$(r)$ $15$ $SI$ unit
$(D)$ Vertical displacement	$(s)$ $20$ $SI$ unit

Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$

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