Respuesta :

elbunt
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]

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