(B) The velocity components are obtained by differentiating the position equations with respect to time $t$.$v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5

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(B) The velocity components are obtained by differentiating the position equations with respect to time $t$.$v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) = 8 - 10t \ m/s$$v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) = 6 \ m/s$At the time of projection,$t = 0$.Substituting $t = 0$ into the velocity components:$v_{y0} = 8 - 10(0) = 8 \ m/s$$v_{x0} = 6 \ m/s$The angle of projection $\theta$ is given by $\tan \theta = \frac{v_{y0}}{v_{x0}}$.$\tan \theta = \frac{8}{6} = \frac{4}{3}$Therefore,$\theta = \tan^{-1}(4/3)$.

Class 11 Physics 2 Motion in a Plane Horizontal Projectile Motion

Medium

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The angle with the horizontal at which the projectile was projected is:

A
$\tan^{-1}(3/4)$
B
$\tan^{-1}(4/3)$
C
$\sin^{-1}(3/4)$
D
Not obtainable from the given data

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2 Motion in a Plane

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Horizontal Projectile Motion

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].

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$A$ body is thrown at an angle of $30^{\circ}$ to the horizontal with a velocity of $30\; m/s$. After $1\; s$,its velocity will be (in $m/s$) $\left(g=10\; m/s^{2}\right)$

Medium

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Match the columns:

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

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

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

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

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

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

Medium

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The initial speed of a projectile fired from the ground is $u$. At the highest point during its motion,the speed of the projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:

MediumJEE MAIN 2023

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Define a projectile and derive the equation of its trajectory: $y = (\tan \theta_0)x - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$.

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The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The angle with the horizontal at which the projectile was projected is:

Similar Questions

$A$ ball rolls off the top of a stairway with horizontal velocity $u$. The steps are $0.1 \ m$ high and $0.1 \ m$ wide. The minimum velocity $u$ with which the ball just hits the $5^{\text{th}}$ step of the stairway will be $\sqrt{x} \ ms^{-1}$ where $x=$ . . . . . . [use $g=10 \ ms^{-2}$].

$A$ body is thrown at an angle of $30^{\circ}$ to the horizontal with a velocity of $30\; m/s$. After $1\; s$,its velocity will be (in $m/s$) $\left(g=10\; m/s^{2}\right)$

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

The initial speed of a projectile fired from the ground is $u$. At the highest point during its motion,the speed of the projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:

Define a projectile and derive the equation of its trajectory: $y = (\tan \theta_0)x - \frac{g}{2(v_0 \cos \theta_0)^2}x^2$.

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Match the columns:

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

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

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

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

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