# A body is thrown up with an initial velocity u, and covers a maximum height of h, then find the value of h.

2
by parasg902

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by parasg902

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2gs = v²-u² g = -9.8 v=0

2×-9.8 ×s =0 - u²

=-u²

-196*s=-u²

s=-u²/-196

=u²/196

h=u²/196

2×-9.8 ×s =0 - u²

=-u²

-196*s=-u²

s=-u²/-196

=u²/196

h=u²/196

-9.8

2x(9.8)x s=0-u power2

=-u power2

-198 x 5 =-u power2

s =- u power 2\ - 196

=u power2 \196

h=u power 2\ 196