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JEE Main 201322 AprilOnline

Exam: JEE Main
Year: 2013
Date: 22 April
Mode: Online
Subject: Physics
Chapter: Motion in Two Dimensions

JEE Main 2013 · 22 April · Online · Physics · Motion in Two Dimensions

A ball projected from ground at an angle of 45^ just clears a wall...

A ball projected from ground at an angle of

45^{\circ}

just clears a wall in front. If point of projection is

4 \mathrm{~m}

from the foot of wall and ball strikes the ground at a distance of

6 \mathrm{~m}

on the other side of the wall, the height of the wall is : A ball projected from ground at an angle of 45^ just clears a wall in front. If point of projection is 4 ~m from the

A ball projected from ground at an angle of 45^ just clears a wall in front. If point of projection is 4 ~m from the

A ball projected from ground at an angle of 45^ just clears a wall in front. If point of projection is 4 ~m from the

Options

A $4.4 \mathrm{~m}$
B $2.4 \mathrm{~m}$
C $3.6 \mathrm{~m}$
D $1.6 \mathrm{~m}$

Correct answer

B. $2.4 \mathrm{~m}$

Step-by-step solution

A ball projected from ground at an angle of 45^ just clears a wall in front. If point of projection is 4 ~m from the

As ball is projected at an angle

45^{{}^{\circ}}

to the horizontal therefore Range

=4 \mathrm{H} or 10=4 \mathrm{H} \Rightarrow \mathrm{H}=\frac{10}{4}=2.5 \mathrm{~m}(\because

Range

=4 \mathrm{~m}+6 \mathrm{~m}=10 \mathrm{~m})

Maximum height,

\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\therefore \mathrm{u}^2=\frac{\mathrm{H} \times 2 \mathrm{~g}}{\sin ^2 \theta}=\frac{2.5 \times 2 \times 10}{\left(\frac{1}{\sqrt{2}}\right)^2}

=100

or,

\mathrm{u}=\sqrt{100}=10 \mathrm{~ms}^{-1}

Height of wall PA

\begin{aligned}& =\mathrm{OA} \tan \theta-\frac{1}{2} \frac{\mathrm{g}(\mathrm{OA})^2}{\mathrm{u}^2 \cos ^2 \theta}\\ & =4-\frac{1}{2} \times \frac{10 \times 16}{10 \times 10 \times \frac{1}{\sqrt{n}} \times \frac{1}{\sqrt{0}}}=2.4 \mathrm{~m} \end{aligned}

$

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