A stone A is projected vertically upwards with a velocity of 39.2 m/s. After one second another stone B
is projected vertically upward from the same point with a velocity of 36.26m/s. Find when and where
they meet? Also find the direction of their motion when they meet
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Answer:The first stone when reaches maximum height, it's velocity is zero. Taking initial velocity as 2 m/s and g as negative 10 m/s^2… and using v=u+gt, it works out that it will reach maximum height in 0.2 seconds. Same time it will take to descend from the point of maximum height to ground. Hence its total time of flight in air is 0.4 s. The second stone is thrown 2 s after the first. Hence they will never meet in air. They meet on ground after first stone arrives back that is 0.4 s after it is thrown up …
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