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Byju's Answer

Standard XII

Physics

Horizontal and Vertical Components of Projectile Motion

A body projec...

Question

A body projected at an angle $45^{o}$ with horizontal has range $16 m$ . It explodes into two parts of equal masses at the highest point One of the parts falls downwards at the point of explosion. At what distance from the point of throw, the other will fall?
( $g = 10 m s^{- 2}$ )

8 m

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16 m

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24 m

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32 m

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Solution

The correct option is C $24 m$

$\begin{matrix} x c m = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} 16 = \frac{m / 2 (8) + m / 2 x_{2}}{(m / 2) + (m / 2)} 16 = \frac{m / 2 (8 + x_{2})}{m / 2 (1 + 1)} 32 = 8 + x_{2} x_{2} = 24 m \end{matrix}$

Hence, Option $C$ is the correct answer.

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A body projected at an angle 45o with horizontal has range 16m. It explodes into two parts of equal masses at the highest point One of the parts falls downwards at the point of explosion. At what distance from the point of throw, the other will fall?(g=10ms−2)

The correct option is C 24mxcm=m1x1+m2x2m1+m216=m/2(8)+m/2x2(m/2)+(m/2)16=m/2(8+x2)m/2(1+1)32=8+x2x2=24mHence, Option C is the correct answer.

A body projected at an angle $45^{o}$ with horizontal has range $16 m$ . It explodes into two parts of equal masses at the highest point One of the parts falls downwards at the point of explosion. At what distance from the point of throw, the other will fall?
( $g = 10 m s^{- 2}$ )

The correct option is C $24 m$

$\begin{matrix} x c m = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} 16 = \frac{m / 2 (8) + m / 2 x_{2}}{(m / 2) + (m / 2)} 16 = \frac{m / 2 (8 + x_{2})}{m / 2 (1 + 1)} 32 = 8 + x_{2} x_{2} = 24 m \end{matrix}$

Hence, Option $C$ is the correct answer.