A ball released from certain height travels
10m in 1.Osec) The distance traveled by
ball after 10 seconds will be
40
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Explanation:
Time of flight =g2Hs
Distance traveled in last second =H-distance traveled in =(g2H−1)s
traveled in =(g2H−1)s
Distance traveld in (g2H−1)s=21g(g2H−1)2
=2g[g2H+1−2g2H]
=H+2g−gg2H
=H+2g−g2Hg
15=H−(H+2g−2gH
=2gH−2g=
traveled in =(g2H−1)s
Distance traveld in (g2H−1)s=21g(g2H−1)2
=2g[g2H+1−2g2H]
=H+2g−gg2H
=H+2g−g2Hg
15=H−(H+2g−2gH
=2gH−2g=
traveled in =(g2H−1)s
Distance traveld in (g2H−1)s=21g(g2H−1)2
=2g[g2H+1−2g2H]
=H+2g−gg2H
=H+2g−g2Hg
15=H−(H+2g−2gH
=2gH−2g=
traveled in =(g2H−1)s
Distance traveld in (g2H−1)s=21g(g2H−1)2
=2g[g2H+1−2g2H]
=H+2g−gg2H
=H+2g−g2Hg
15=H−(H+2g−2gH
=2gH−2g=
traveled in =(g2H−1)s
Distance traveld in (g2H−1)s=21g(g2H−1)2
=2g[g2H+1−2g2H]
=H+2g−gg2H
=H+2g−g2Hg
15=H−(H+2g−2gH
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