A stone is thrown in a vertically upward direction with a velocity of $6 m s^{- 1}$ . If the acceleration of the stone during its motion is $12 m s^{- 2}$ in the downward direction, what will be the height attained by the stone?

1.5 m

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1.2 m

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0.25 m

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0.5 m

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Solution

The correct option is A $1.5 m$
u = $6 m s^{- 1}$
v = 0
a = $- 12 m s^{- 2}$
s = ?
t = ?
From the first equation of motion,
v = u + at
0 = 6 + (-12)t
-6 = -12t
t = $\frac{6}{12}$ = 0.5 s
From the third equation of motion,
$v^{2}$ - $u^{2}$ = 2as
$(0)^{2}$ - $(6)^{2}$ = 2 $\times$ (-12) $\times$ s
- 36 = - 24 $\times$ s
s = $\frac{36}{24}$ = 1.5 m

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A stone is thrown in a vertically upward direction with a velocity of 6 ms−1. If the acceleration of the stone during its motion is 12 ms−2 in the downward direction, what will be the height attained by the stone?

The correct option is A 1.5 mu = 6 ms−1 v = 0 a = −12 ms−2 s = ? t = ? From the first equation of motion, v = u + at 0 = 6 + (-12)t -6 = -12t t = 612 = 0.5 s From the third equation of motion, v2 - u2 = 2as (0)2 - (6)2 = 2 × (-12) × s - 36 = - 24 × s s = 3624 = 1.5 m

A stone is thrown in a vertically upward direction with a velocity of $6 m s^{- 1}$ . If the acceleration of the stone during its motion is $12 m s^{- 2}$ in the downward direction, what will be the height attained by the stone?