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Standard XII

Physics

Variation of G Due to the Rotation of Earth

A man riding ...

Question

A man riding on a flat car moving with $10 m s^{- 1}$ . He attempts to throw a ball through a stationary hoop $5 m$ above his hand such that the ball moves horizontally through the hoop. He throws the ball with $12 m s^{- 1}$ with respect to himself. Find the horizontal distance from where he throws the ball

15 m

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14.2 m

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16.7 m

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18.2 m

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Solution

The correct option is C $16.7 m$
$h_{m a x} = 5 = \frac{u_{y}^{2}}{2 g}$
$u_{y} = 10$
$u_{x} = \sqrt{12^{2} - 10^{2}} = \sqrt{44}$
$v_{x} = 10 + \sqrt{44}$
$\frac{T}{2} = \frac{u_{y}}{g} = 1 s e c$
$x = v_{x} . \frac{T}{2} = 1 \times (10 + \sqrt{44}) = 16.7 m$

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A man riding on a flat car moving with 10 ms−1. He attempts to throw a ball through a stationary hoop 5 m above his hand such that the ball moves horizontally through the hoop. He throws the ball with 12 ms−1 with respect to himself. Find the horizontal distance from where he throws the ball

The correct option is C 16.7 mhmax=5=u2y2guy=10ux=√122−102=√44vx=10+√44 T2=uyg=1secx=vx.T2=1×(10+√44)=16.7m

The correct option is C $16.7 m$
$h_{m a x} = 5 = \frac{u_{y}^{2}}{2 g}$
$u_{y} = 10$
$u_{x} = \sqrt{12^{2} - 10^{2}} = \sqrt{44}$
$v_{x} = 10 + \sqrt{44}$
$\frac{T}{2} = \frac{u_{y}}{g} = 1 s e c$
$x = v_{x} . \frac{T}{2} = 1 \times (10 + \sqrt{44}) = 16.7 m$