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(C) The initial velocity is $u = 100\,m/s$ at an angle $	heta = 53^{\circ}$.Horizontal component: $u_x = u \cos 53^{\circ} = 100 	imes 0.6 = 60\,m/s

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(C) The initial velocity is $u = 100\,m/s$ at an angle $	heta = 53^{\circ}$.Horizontal component: $u_x = u \cos 53^{\circ} = 100 	imes 0.6 = 60\,m/s$.Vertical component: $u_y = u \sin 53^{\circ} = 100 	imes 0.8 = 80\,m/s$.The horizontal component $v_x$ remains constant at $60\,m/s$ throughout the motion.For the velocity vector to make an angle of $45^{\circ}$ with the horizontal,the vertical component $v_y$ must satisfy $	an 45^{\circ} = |v_y / v_x|$,which implies $|v_y| = v_x = 60\,m/s$.Case $1$: $v_y = +60\,m/s$ (during ascent).Using $v_y = u_y + a_y t_1$ with $a_y = -10\,m/s^2$:$60 = 80 - 10 t_1 \implies 10 t_1 = 20 \implies t_1 = 2\,s$.Case $2$: $v_y = -60\,m/s$ (during descent).$-60 = 80 - 10 t_2 \implies 10 t_2 = 140 \implies t_2 = 14\,s$.Thus,both $2\,s$ and $14\,s$ are possible times.

$A$ particle is projected from the horizontal making an angle of $53^{\circ}$ with an initial velocity of $100\,m/s$. The time taken by the particle to make an angle of $45^{\circ}$ with the horizontal is $.........\,s$.

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