Categories > differentiation > Application of Derivatives > The equations of motion of two stones thrown vertically upwards simultaneously are $s=19.6\,t-4.9\,{{t}^{2}}$ and $s=9.8\,t-4.9\,{{t}^{2}}$ respectively and the maximum height attained by the first one is h. When the height of the first stone is maximum, the height of the second stone will be

differentiation

Question: The equations of motion of two stones thrown vertically upwards simultaneously are $s=19.6\,t-4.9\,{{t}^{2}}$ and $s=9.8\,t-4.9\,{{t}^{2}}$ respectively and the maximum height attained by the first one is h. When the height of the first stone is maximum, the height of the second stone will be


1) h/3
2) 2h
3) h
4) 0
Solution: Explanation: No Explanation
Application of Derivatives

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