Scientific measurement - 2 | Chemistry and the Scientific Method

At t = 0, the height = 0 m, so E_p0 = mgh = 0 J

At t = t_f  (final time), the ball reaches the top of its trajectory 
⇒ the velocity = 0 m.s^-1 
⇒ E_{k tf} = ½ mv² = 0 J.
At this time we want h > 30 m, so E_{p tf} > 30 x m x g

Law of conservation of energy:
E_total = cste = E_k + E_p = ½ mv² + mgh
and E_total (t = 0) = E_total (t = t_f)

E_{k 0} = E_{p tf} > 30 x m x g
½ x m x v₀² > 30 x m x g
v₀ > (2 x 30 x g)^0.5

v₀ > 24 m.s^-1