a stone is thrown vertically upwards from the roof of a building with a velocity of 30m/s, another stone is dropped 4 seconds after the first is thrown up. Find when and where the two stones meet.
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hi thank you for asking the question
when the stones are thrown upward and another thrown downward there velocities are known so by using the formula for stone 1 y1=1/2gt^2 and similar formula for the 2nd stone we can calculate the distance covered by them in unit time,comparing the equation of both we can get the result
25m will be the distance when both the stones will meet
i hope my answer was helpful
when the stones are thrown upward and another thrown downward there velocities are known so by using the formula for stone 1 y1=1/2gt^2 and similar formula for the 2nd stone we can calculate the distance covered by them in unit time,comparing the equation of both we can get the result
25m will be the distance when both the stones will meet
i hope my answer was helpful
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