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(A) The horizontal position is given by $x = 36t$. The horizontal velocity is $v_x = \frac{dx}{dt} = 36 \, m/s$.The vertical position is given by $2y

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{4}{5}\right)$

Vedclass · Answer

(A) The horizontal position is given by $x = 36t$. The horizontal velocity is $v_x = \frac{dx}{dt} = 36 \, m/s$.The vertical position is given by $2y = 96t - 9.8t^2$,which simplifies to $y = 48t - 4.9t^2$. The vertical velocity is $v_y = \frac{dy}{dt} = 48 - 9.8t$.At the time of projection $(t = 0)$,the initial horizontal velocity is $u_x = 36 \, m/s$ and the initial vertical velocity is $u_y = 48 \, m/s$.The angle of projection $	heta$ is given by $	an 	heta = \frac{u_y}{u_x} = \frac{48}{36} = \frac{4}{3}$.Using the trigonometric identity for a right-angled triangle with opposite side $4$ and adjacent side $3$,the hypotenuse is $\sqrt{4^2 + 3^2} = 5$. Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{4}{5}$.Thus,$	heta = \sin^{-1}\left(\frac{4}{5}ight)$.

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

The equations of motion of a projectile are given by $x = 36t \, m$ and $2y = 96t - 9.8t^2 \, m$. The angle of projection is

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The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$

Match the columns:

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

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

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

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

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

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

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