A boy standing on the surface of the Moon throws a ball of mass $1 kg$ vertically upwards with an initial speed of $20 m/s$ . What is the maximum height attained by the ball? (Given: Weight of the object on Moon $= \frac{1}{6}$ weight of the object on Earth; Acceleration due to gravity on Earth $= 10 {m/s}^{2}$

5 m

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20 m

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120 m

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60 m

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Solution

The correct option is C $120 m$
Acceleration due to the gravity of Moon $= \frac{1}{6}$ (Acceleration due to gravity of Earth)
$g_{moon} = \frac{1}{6} g$
By using the third equation of motion,
$v^{2} - u^{2} = 2 a s$
$(0)^{2} - (20)^{2} = 2 (- \frac{1}{6} \times 10) s$
$s = 120 m$

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A boy standing on the surface of the Moon throws a ball of mass 1 kg vertically upwards with an initial speed of 20 m/s. What is the maximum height attained by the ball? (Given: Weight of the object on Moon =16 weight of the object on Earth; Acceleration due to gravity on Earth =10 m/s2

The correct option is C 120 mAcceleration due to the gravity of Moon =16 (Acceleration due to gravity of Earth) gmoon=16 g By using the third equation of motion, v2−u2=2as (0)2−(20)2=2(−16×10) s s=120 m

The correct option is C $120 m$
Acceleration due to the gravity of Moon $= \frac{1}{6}$ (Acceleration due to gravity of Earth)
$g_{moon} = \frac{1}{6} g$
By using the third equation of motion,
$v^{2} - u^{2} = 2 a s$
$(0)^{2} - (20)^{2} = 2 (- \frac{1}{6} \times 10) s$
$s = 120 m$