JAMB - Physics (1986 - No. 6)
A ball is thrown vertically into the air with an initial velocity u. What is the greatest height reached?
u2g
\(\frac{3u^2}{2g}\)
\(\frac{u^2}{g}\)
\(\frac{u^2}{2g}\)
Wyjaśnienie
Using V2 = U2 - 2gS
V = 0ms-1 at greatest height
S = \(\frac{u^2}{2g}\)
V = 0ms-1 at greatest height
S = \(\frac{u^2}{2g}\)