Kinematic question!
start with our knowns
X=1 500 000
a=-9.8 m/s^2 because of gravity
t=who cares
startV=?
finalV=0 (this is the minimum, it could go higher but we only want the threshold)
solve for startV!
with the information we have we can only use the fourth kinematic!
[tex]{finalv}^{2} = {startv}^{2} + 2 \times a \times x \\ 0 = {startv}^{2} + 2 \times ( - 9.8) \times (1 \: 500 \: 000) \\ startv = \sqrt{29400000} = 5422 \frac{m}{s} [/tex]
