Question

ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after time. (A)         v2 2gh 1 1 g v (B)         v2 2gh 1 1 g v (C) 1/ 2 v2 2gh 1 g v      (D) 1/ 2 v2 2gh 1 g v     ​

Answer 1

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Answer : vg[1+1+2ghv2−−−−−−√]

h=−vt+12gt2

or gt2−2vt−h=0

t=−(−2v)±4v2+4gh−−−−−−−√2g

=2v±2v2+gh−−−−−√2g

=vg±[v2+2gh]1/2g

=vg[1+1+2ghv2−−−−−−√]

Now retain only the positive sign.

hope this helps ✌️❤️✌️

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Answer 2

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Answer : vg[1+1+2ghv2−−−−−−√]

h=−vt+12gt2

or gt2−2vt−h=0

t=−(−2v)±4v2+4gh−−−−−−−√2g

=2v±2v2+gh−−−−−√2g

=vg±[v2+2gh]1/2g

=vg[1+1+2ghv2−−−−−−√]

Now retain only the positive sign.