Question

 vg[1+1+2ghv2−−−−−−√]vg[1+1+2ghv2]

h=−vt+12h=−vt+12gt2gt2

or gt2−2vt−h=0gt2−2vt−h=0

t=−(−2v)±4v2+4gh−−−−−−−√2gt=−(−2v)±4v2+4gh2g

=2v±2v2+gh−−−−−√2g=2v±2v2+gh2g

=vg±[v2+2gh]1/2g=vg±[v2+2gh]1/2g

=vg[1+1+2ghv2−−−−−−√]=vg[1+1+2ghv2]

​

Answer 1

Explanation:

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.

Answer 2

Answer:

This is not a question. This is a Answer.

Hope it Helps !!!!