Posted by monnie on Wednesday, October 6, 2010 at 11:27pm.
Kinematics Equations:
Vx1 = Vx0 + axΔt
(Vx1)^2 = (Vx0)^2 + 2axΔx
Δx = Vx0Δt + ½ ax(Δt^2)
Using the second kinematics equation, we can solve for Δx.
(Vx1)^2 = (Vx0)^2 + 2axΔx
Plug in what you know:
(45 m/s)^2 = (0 m/s) + 2(9.8m/s^2)(? m)
2025 m^2/s^2 = (19.6 m/s^2)(? m)
(? m) = (2025 m^2/s^2)/(19.6 m/s^2)
? m = 103.3163265
(sig figs) = 100 m
Δx = 100m
-----------------------------------
Using the first kinematics equation, we can solve for Δt.
Vx1 = Vx0 + axΔt
Plug in what you know:
Vx1 = Vx0 + axΔt
45 m/s = 0 m/s + (9.8 m/s^2)(Δt)
Δt = (45 m/s)/(9.8 m/s^2)
Δt = 4.591836735
(sig figs) =
Δt = 4.6 s
Let me know if that helped you!
yes it did help and i get it now thank you
shouldn't acceleration be -9.8 and not +9.8 since it freefalling in the negative direction?
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