A cricket ball is thrown up with a speed of $19.6 m s^{- 1}$ . The maximum height it can reach is

9.8 m

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19.6 m

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29.4 m

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39.2 m

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Solution

The correct option is B $19.6 m$
Given,
$u = 19.6 m / s$
$v = 0 m / s$
$g = 9.8 m / s^{2}$
From 3rd equation of motion,
$2 g h = v^{2} - u^{2}$
$- 2 \times 9.8 \times h = 0^{2} - 19.6 \times 19.6$
$h = \frac{19.6 \times 19.6}{2 \times 9.8} = 19.6 m$
The correct option is B.

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The maximum height which can be reached is:

A body projected upwards with an initial velocity of $49 \frac{m}{s}$ . What will be its final velocity after 3 seconds? [acceleration due to gravity $g = 9.8 \frac{m}{s^{2}}$ ]

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reaches a height of

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g_{m o o n} = \frac{g}{6} m s^{- 2}

)

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)

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A cricket ball is thrown up with a speed of 19.6 ms−1. The maximum height it can reach is

The correct option is B 19.6 mGiven,u=19.6m/sv=0m/sg=9.8m/s2From 3rd equation of motion,2gh=v2−u2−2×9.8×h=02−19.6×19.6h=19.6×19.62×9.8=19.6mThe correct option is B.

A cricket ball is thrown up with a speed of $19.6 m s^{- 1}$ . The maximum height it can reach is

The correct option is B $19.6 m$
Given,
$u = 19.6 m / s$
$v = 0 m / s$
$g = 9.8 m / s^{2}$
From 3rd equation of motion,
$2 g h = v^{2} - u^{2}$
$- 2 \times 9.8 \times h = 0^{2} - 19.6 \times 19.6$
$h = \frac{19.6 \times 19.6}{2 \times 9.8} = 19.6 m$
The correct option is B.