At a construction site a pipe wrench struck the ground with a speed of 45 m/s. (a) From what height was it inadvertently dropped? (b) How long was it falling?
shouldn't acceleration be -9.8 and not +9.8 since it freefalling in the negative direction?
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
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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
To solve this problem, we can use the equations of motion under constant acceleration. The initial velocity (u) of the pipe wrench when it hits the ground is 0 m/s because it was dropped from rest. The final velocity (v) of the pipe wrench is 45 m/s.
(a) We want to find the height (h) from which the pipe wrench was dropped.
We can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (45 m/s)
u = initial velocity (0 m/s)
a = acceleration (acceleration due to gravity, approximately 9.8 m/s^2)
s = displacement (height, in this case)
Rearranging the equation, we have:
s = (v^2 - u^2) / 2a
Substituting the known values:
s = (45^2 - 0^2) / (2 * 9.8)
Simplifying:
s = (2025) / 19.6
s ≈ 103.57m
Therefore, the pipe wrench was dropped from a height of approximately 103.57 meters.
(b) To find the time (t) it took for the pipe wrench to fall, we can use the equation:
v = u + at
Rearranging the equation, we have:
t = (v - u) / a
Substituting the known values:
t = (45 - 0) / 9.8
Simplifying:
t ≈ 4.59s
Therefore, the pipe wrench was falling for approximately 4.59 seconds.