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April 18, 2015

April 18, 2015

Posted by **monnie** on Wednesday, October 6, 2010 at 11:27pm.

- physics -
**Steven**, Thursday, October 7, 2010 at 12:29amKinematics Equations:

Vx1 = Vx0 + axΔt

(Vx1)^2 = (Vx0)^2 + 2axΔx

Δx = Vx0Δt + ˝ 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!

- physics -
**monnie**, Thursday, October 7, 2010 at 12:53amyes it did help and i get it now thank you

- physics -
**maria**, Friday, February 4, 2011 at 3:35pmshouldn't acceleration be -9.8 and not +9.8 since it freefalling in the negative direction?

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