(C) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g}$.For th

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(C) The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$.For maximum range,$\theta = 45^{\circ}$,so $R_{\max} = \frac{u^2}{g}$.For the second object,the range is $R' = \frac{R_{\max}}{2} = \frac{u^2}{2g}$.Equating the range formula for the second object: $\frac{u^2 \sin 2\theta'}{g} = \frac{u^2}{2g}$.This simplifies to $\sin 2\theta' = \frac{1}{2}$.Therefore,$2\theta' = 30^{\circ}$ or $2\theta' = 150^{\circ}$.Solving for $\theta'$,we get $\theta' = 15^{\circ}$ or $\theta' = 75^{\circ}$.

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

Medium

An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$ is maximum. Another object is projected in the air with a horizontal range half of the range of the first object. The initial velocity remains the same in both cases. The value of the angle of projection,at which the second object is projected,will be $.......$ degrees.

JEE MAIN 2022 All JEE MAIN

A
$85$
B
$80$
C
$15$ or $75$
D
$70$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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When a ball is thrown with a velocity of $50 \ m s^{-1}$ at an angle $30^{\circ}$ with the horizontal,it remains in the air for how many seconds? (Take $g = 10 \ m s^{-2}$)

EasyAP EAMCET 2021

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The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

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Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed $(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed $(b)$ $0^o$

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$A$ particle is projected with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in its velocity at the maximum height?

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$A$ 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:

MediumAP EAMCET 2020

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Column-$I$	Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed	$(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed	$(b)$ $0^o$

Similar Questions

When a ball is thrown with a velocity of $50 \ m s^{-1}$ at an angle $30^{\circ}$ with the horizontal,it remains in the air for how many seconds? (Take $g = 10 \ m s^{-2}$)

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

Match Column-$I$ with Column-$II$.Column-$I$Column-$II$$(1)$ Angle of projection for a projectile launched horizontally with constant speed$(a)$ $0$$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed$(b)$ $0^o$

$A$ particle is projected with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in its velocity at the maximum height?

$A$ 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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Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Angle of projection for a projectile launched horizontally with constant speed $(a)$ $0$
$(2)$ Horizontal component of acceleration for a projectile launched horizontally with constant speed $(b)$ $0^o$