(B) Let the angle of projection of particle $B$ with the horizontal be $\alpha$. For particle $B$ to hit particle $A$ moving on the ground,the horizon

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(B) Let the angle of projection of particle $B$ with the horizontal be $\alpha$. For particle $B$ to hit particle $A$ moving on the ground,the horizontal component of the velocity of $B$ must be equal to the velocity of $A$ (since $A$ moves with constant speed $v$ on the horizontal surface). Thus,$v_B \cos \alpha = v$. Given $v_B = \frac{2v}{\sqrt{3}}$,we have $\frac{2v}{\sqrt{3}} \cos \alpha = v$. $\cos \alpha = \frac{\sqrt{3}}{2}$. Therefore,$\alpha = 30^{\circ}$ with the horizontal. The question asks for the angle of projection with the vertical. Angle with vertical = $90^{\circ} - \alpha = 90^{\circ} - 30^{\circ} = 60^{\circ}$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

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Two particles $A$ and $B$ are projected simultaneously from a fixed point on the ground. Particle $A$ is projected on a smooth horizontal surface with speed $v$,while particle $B$ is projected in air with speed $\frac{2v}{\sqrt{3}}$. If particle $B$ hits particle $A$,the angle of projection of $B$ with the vertical is

A
$30^{\circ}$
B
$60^{\circ}$
C
$45^{\circ}$
D
Both $(a)$ and $(b)$

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3-2.Motion in Plane

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Horizontal Projectile Motion

The angular momentum of a projectile projected at an angle $\theta$ with the horizontal with speed $u$ about the point of projection when it is at the highest point of its trajectory is

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An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

DifficultAP EAMCET 2001

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Two spheres are projected at angles $30^{\circ}$ and $45^{\circ}$ with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, $(\sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 30^{\circ} = 0.5)$

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$A$ fighter plane, flying horizontally with a speed of $360 \text{ km/h}$ at an altitude of $500 \text{ m}$, drops a bomb for a target straight ahead of it on the ground. At what approximate distance ahead of the target should the bomb be dropped? Assume that acceleration due to gravity $g = 10 \text{ m/s}^2$. Neglect air drag.

MediumWBJEE 2020

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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 $y = 8t - 5t^2 \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The velocity with which the projectile is projected is (in $\text{ m/s}$)

MediumKCET 2011

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The angular momentum of a projectile projected at an angle $\theta$ with the horizontal with speed $u$ about the point of projection when it is at the highest point of its trajectory is

An object is projected with a velocity of $20 \ m/s$ making an angle of $45^{\circ}$ with the horizontal. The equation for the trajectory is $h = Ax - Bx^2$, where $h$ is the height, $x$ is the horizontal distance, and $A$ and $B$ are constants. The ratio $A:B$ is $(g = 10 \ m/s^2)$.

Two spheres are projected at angles $30^{\circ}$ and $45^{\circ}$ with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, $(\sin 45^{\circ} = \frac{1}{\sqrt{2}}, \sin 30^{\circ} = 0.5)$

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 $y = 8t - 5t^2 \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The velocity with which the projectile is projected is (in $\text{ m/s}$)

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