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An object is thrown vertically upward with a speed of $30 \mathrm{~m} / \mathrm{s}$. The velocity of the object half-a-second before it reaches the maximum height is
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Verified Answer
The correct answer is:
$4.9 \mathrm{~m} / \mathrm{s}$
According to the question,
Velocity of an object half a second before maximum height
= Velocity of an object half a second after maximum height (return journey)
$$
=0+g t=0+9.8 \times \frac{1}{2}=4.9 \mathrm{~m} / \mathrm{s}
$$
Velocity of an object half a second before maximum height
= Velocity of an object half a second after maximum height (return journey)
$$
=0+g t=0+9.8 \times \frac{1}{2}=4.9 \mathrm{~m} / \mathrm{s}
$$
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