A boy playing on roof of a 10, m high building throws a ball with a speed of 10, m/s at an angle 30^

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3-2.Motion in Plane

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A boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\, m/s$ at an angle $30^o$ with the horizontal. ........ $m$ far from the throwing point will the ball be at the height of $10\, m$ from the ground . $(g \,= \,10 m/s^2, \,sin \,30^o \,= \,\frac{1}{2}$, $\cos \,{30^o}\, = \,\frac{{\sqrt 3 }}{2}$)

A$5.20$

B$4.33$

C$2.60$

D$8.66$

(AIEEE-2003)

Solution

$ \mathrm{R} =\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}=\frac{(10)^{2} \sin 2 \times 30^{\circ}}{10}$
$=\frac{10 \sqrt{3}}{2}$
$=5 \times 1.732=8.66 \mathrm{m} $

Standard 11

Physics

The initial velocity of a particle of mass $2\,kg$ is $(4 \hat{ i }+4 \hat{ j })\,m / s$. A constant force of $-20 \hat{ j }\,N$ is applied on the particle. Initially, the particle was at $(0,0)$. Find the $x$-coordinate of the point where its $y$-coordinate is again zero.$……….\,m$

medium

Two balls are projected with the same velocity but with different angles with the horizontal. Their ranges are equal. If the angle of projection of one is $30^{\circ}$ and its maximum height is $h$, then the maximum height of other will be

hard

(KVPY-2020)

For an object projected from ground with speed $u$ horizontal range is two times the maximum height attained by it. The horizontal range of object is ……….

hard

Four bodies $P, Q, R$ and $S$ are projected with equal velocities having angles of projection $15^o , 30^o , 45^o $ and $60^o $ with the horizontal respectively. The body having shortest range is

easy

Column $-I$
Angle of projection Column $-II$

$A.$ $\theta \, = \,{45^o}$ $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$

$B.$ $\theta \, = \,{60^o}$ $2.$ $\frac{{g{T^2}}}{R} = 8$

$C.$ $\theta \, = \,{30^o}$ $3.$ $\frac{R}{H} = 4\sqrt 3 $

$D.$ $\theta \, = \,{\tan ^{ – 1}}\,4$ $4.$ $\frac{R}{H} = 4$

$K_i :$ initial kinetic energy
$K_h :$ kinetic energy at the highest point

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Column $-I$ Angle of projection	Column $-II$
$A.$ $\theta \, = \,{45^o}$	$1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$
$B.$ $\theta \, = \,{60^o}$	$2.$ $\frac{{g{T^2}}}{R} = 8$
$C.$ $\theta \, = \,{30^o}$	$3.$ $\frac{R}{H} = 4\sqrt 3 $
$D.$ $\theta \, = \,{\tan ^{ – 1}}\,4$	$4.$ $\frac{R}{H} = 4$

Solution

Similar Questions

The initial velocity of a particle of mass $2\,kg$ is $(4 \hat{ i }+4 \hat{ j })\,m / s$. A constant force of $-20 \hat{ j }\,N$ is applied on the particle. Initially, the particle was at $(0,0)$. Find the $x$-coordinate of the point where its $y$-coordinate is again zero.$……….\,m$

Two balls are projected with the same velocity but with different angles with the horizontal. Their ranges are equal. If the angle of projection of one is $30^{\circ}$ and its maximum height is $h$, then the maximum height of other will be

For an object projected from ground with speed $u$ horizontal range is two times the maximum height attained by it. The horizontal range of object is ……….

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