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Motion in a Plane

Section: Physics

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Question : 28 of 28

Marks: +1, -0

A ball is projected at an angle of

45^{\circ}

with the horizontal. It passes through a wall of height '

h

' at a horizontal distance

d_1

from the point of projection and strikes the ground at a distance

d_1+d_2

from the point of projection, then '

h

[AP EAPCET 18 May 2024 Shift 1]

$h=\frac{2 d_1 d_2}{d_1+d_2}$
$h=\frac{d_1 d_2}{d_1+d_2}$
$h=\frac{\sqrt{2} d_1 d_2}{d_1+d_2}$
$h=\frac{d_1 d_2}{2 (d_1+d_2)}$

Solution:

Angle of projection

=45^{\circ}

height of wall

=h

We know that.

The equation of trajectory of a projectale,

y=x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}

y=x \tan \theta \left(1-\frac{g x}{2 u^2 \cos^2 \theta \cdot \tan \theta}\right)

y=x \tan \theta \left(1-\frac{g x}{2 u^2 \cos^2 \theta \cdot \frac{\sin \theta}{\cos \theta}}\right)

y=x \tan \theta \left(1-\frac{x}{R}\right) \{\because R=\frac{u^2 \sin 2\theta}{g}\}

Here,

R=d_1+d_2, x=d_1

and

y=h

So,

h=d_1 \tan 45^{\circ} \left(1-\frac{d_1}{d_1+d_2}\right)

h=d_1 \left(\frac{d_2}{d_1+d_2}\right)

h=\frac{d_1 d_2}{d_1+d_2}

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