Question

a particle projected with a velocity 49m/s at one angle 30 to the horizontal. calculate the maximum height time of flight and horizontal range​

Answer 1

Topic :- Motion in Plane

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✏ As we know that acceleration in Horizontal direction is always zero $\sf a_x = 0$. Hence, the velocity in the horizontal direction is constant.

✏Initial velocity in the vertical direction is given as $\sf U_y = 49 \:m/s$.

✏The angle of projection is given as 30°.

✏We need to find the Maximum Height (H) attained by the particle, Time of flight (T) and Horizontal Range (R).

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✪ CALCULATION ✪

First we will find the maximum height attained by the body :

$\longrightarrow\:\:\sf Maximum \: Height_{(attained \: by \: the \: body)} = \dfrac{(u _{y})^{2} }{2g}$

$\longrightarrow\:\:\sf Maximum \: Height_{(attained \: by \: the \: body)} = \dfrac{u^2 \sin^2 (\theta)}{2g}$

$\longrightarrow\:\:\sf Maximum \: Height_{(attained \: by \: the \: body)} = \dfrac{(49)^2 \sin^2 ( {30}^{ \circ} )}{2 \times 10}$

$\longrightarrow\:\:\sf Maximum \: Height_{(attained \: by \: the \: body)} = \dfrac{2401 \times ( \frac{1}{2} )^{2} }{20}$

$\longrightarrow\:\:\sf Maximum \: Height_{(attained \: by \: the \: body)} = \dfrac{2401 \times \frac{1}{4} }{20}$

$\longrightarrow\:\:\sf Maximum \: Height_{(attained \: by \: the \: body)} = \dfrac{600.25 }{20}$

$\longrightarrow\:\: \underline{ \underline{\sf Maximum \: Height_{(attained \: by \: the \: body)} = 30.0125 \: m}}$

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Now, let's find Time of flight (T) :

$\longrightarrow\:\:\sf Time \: of \: Flight= \dfrac{2u_y}{g}$

$\longrightarrow\:\:\sf Time \: of \: Flight= \dfrac{2u \sin( \theta) }{g}$

$\longrightarrow\:\:\sf Time \: of \: Flight= \dfrac{2 \times 49 \sin( {30}^{ \circ} ) }{10}$

$\longrightarrow\:\:\sf Time \: of \: Flight= \dfrac{90 \sin( {30}^{ \circ} ) }{10}$

$\longrightarrow\:\:\sf Time \: of \: Flight=9 \sin( {30}^{ \circ} )$

$\longrightarrow\:\:\sf Time \: of \: Flight=9 \times \dfrac{1}{2}$

$\longrightarrow\:\: \underline{ \underline{\sf Time \: of \: Flight=4.5 \: s}}$

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Finding the Horizontal Range (R) :

$\longrightarrow\:\:\sf Horizontal \: Range = \dfrac{u^2 \sin2( \theta) }{g}$

$\longrightarrow\:\:\sf Horizontal \: Range = \dfrac{(49)^{2} \sin2( {30}^{ \circ} ) }{10}$

$\longrightarrow\:\:\sf Horizontal \: Range = \dfrac{2401 \times 2 \times \frac{1}{2} }{10}$

$\longrightarrow\:\:\sf Horizontal \: Range = \dfrac{2401}{10}$

$\longrightarrow\:\: \underline{ \underline{\sf Horizontal \: Range = 240.1 \: m}}$

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