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A ball thrown vartically upwords with the speed 9.8m/s how high will the ball rise and how long will it rise |
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Answer» Let X be the meeting POINT. Both BALLS would have traveled same time to get to meeting point, say 't' secs. Let H 1 be height traveled by ball falling down to X. Let h 2 be height traveled by ball going up to X and v be its velocity. h 1+h 2 =100m v 2=2gH=2×9.8×100=1960 Hencev= 1960 m/s h 1 = 2 1 GT 2 For the ball travelling up, h 2 =vt− 21 gt 2 =vt−h 1 h 1 +h +2 =vt Hence vt=100 t= v 100 =1960 100 secs. h 1 = 2 1 gt 2 = 21 ×9.8× 1960 100×100 = 4 100 =25m h= 2 =100−25=75mExplanation: |
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