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(A) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal.At $t = 3 \,s$ (since $1 \,s + 2 \,s = 3 \,s$), the projectile is moving

Vedclass · Accepted Answer

$10 \sqrt{13} \,ms^{-1}$

Vedclass · Answer

(A) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal.At $t = 3 \,s$ (since $1 \,s + 2 \,s = 3 \,s$), the projectile is moving horizontally, which means the vertical component of velocity is zero at this point.$v_y = u \sin 	heta - gt = 0 \Rightarrow u \sin 	heta = g 	imes 3 = 10 	imes 3 = 30 \,ms^{-1}$.At $t = 1 \,s$, the direction is inclined at $45^{\circ}$ to the horizontal, so $	an 45^{\circ} = \frac{v_y}{v_x} = 1$.Thus, $v_y = v_x \Rightarrow u \sin 	heta - g(1) = u \cos 	heta$.Substituting $u \sin 	heta = 30$ and $g = 10$, we get $30 - 10 = u \cos 	heta \Rightarrow u \cos 	heta = 20 \,ms^{-1}$.The magnitude of the initial velocity is $u = \sqrt{(u \cos 	heta)^2 + (u \sin 	heta)^2} = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} = 10 \sqrt{13} \,ms^{-1}$.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

One second after projection, a projectile is travelling in a direction inclined at $45^{\circ}$ to the horizontal. After two more seconds, it is travelling horizontally. Then the magnitude of the initial velocity of the projectile is $(g = 10 \,ms^{-2})$

Similar Questions

Match the columns:

Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$)

$A. 1$ $1. 60^o$

$B. 4$ $2. 30^o$

$C. 4\sqrt{3}$ $3. 45^o$

$D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$

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