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(C) The time taken by the first stone to reach the maximum height is given by $t = \frac{u}{g} = \frac{50}{10} = 5 \,s$.Since the second stone is thro

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(C) The time taken by the first stone to reach the maximum height is given by $t = \frac{u}{g} = \frac{50}{10} = 5 \,s$.Since the second stone is thrown downwards, its initial velocity is $u = -50 \,ms^{-1}$ (taking upward as positive and downward as negative) and acceleration is $a = -g = -10 \,ms^{-2}$.Alternatively, using the downward direction as positive: $u = 50 \,ms^{-1}$ and $a = 10 \,ms^{-2}$.Using the first equation of motion $v = u + at$ for the second stone at $t = 5 \,s$:$v = 50 + (10 	imes 5) = 50 + 50 = 100 \,ms^{-1}$.

$A$ stone is thrown upwards with a velocity $50 \,ms^{-1}$. Another stone is simultaneously thrown downwards from the same location with a velocity $50 \,ms^{-1}$. When the first stone is at the highest point, the velocity of the second stone is $...... \,ms^{-1}$ (Take $g = 10 \,ms^{-2}$).

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