A stone is thrown upwards with a velocity $50 m / s$ . Another stone is simultaneously thrown downwards from the same location with a velocity $50 m / s$ . When the first stone is at the highest point, the velocity of the second stone is (Take $g = 10 m / s^{2}$ ):

100 m / s

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zero

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150 m / s

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50 m / s

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Solution

The correct option is A $100 m / s$
Given that both the stones are thrown from same point.
Taking the upward direction as positive and downward as negative.

For first stone: When this stone reaches at highest point, its final velocity becomes zero i.e. $v = 0$ and $g = - g$ , $u = 50 m / s$ , time at highest point = $t_{h}$
From first equation of motion $v = u + a t$
Substituting the values, we get $0 = 50 - g t_{h}$ ⇒ $t_{h} = 5$ $s (∵ g = 10 m / s^{2})$

For second stone: $u = - 50 m / s$ ,
From first equation of motion
$v = u + a t$
Substituting the values, we get,
$v = - 50 - 10 \times t_{h}$
$v = - 100 m / s$
$- v e$ sign shows that $v$ will be downward.

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Standard VI Physics

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A stone is thrown upwards with a velocity 50 m/s. Another stone is simultaneously thrown downwards from the same location with a velocity 50 m/s. When the first stone is at the highest point, the velocity of the second stone is (Take g=10m/s2):

A stone is thrown upwards with a velocity $50 m / s$ . Another stone is simultaneously thrown downwards from the same location with a velocity $50 m / s$ . When the first stone is at the highest point, the velocity of the second stone is (Take $g = 10 m / s^{2}$ ):