- #1
rsala
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nevermind, I've solved it, thanks anyway
hello, I've tried this simple problem a few times and mastering physics keeps taking points off =(.
ive never used latex so bare with me
A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground View Figure . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.
Compute the velocity of the sandbag at a time 1.15 s after its release.
x = X[tex]_{0}[/tex] +V[tex]_{0}[/tex]T + .5aT[tex]^{2}[/tex]
V[tex]^{2}[/tex] = V[tex]^{2}_{0}[/tex] + 2a(x - x[tex]_{0}[/tex])
I assume that the initial velocity of the sandbag once it is released is 5ms since that was the velocity of the balloon.
first i found the position of the sandbag at 1.15.
x = 40 + 5(1.15) + .5(-9.81)(1.15)[tex]^{2}[/tex]
x = 39.3
up to this point i am 100% sure i am correct.
next i inserted this information into the velocity formula.
V[tex]^{2}[/tex] = V[tex]^{2}_{0}[/tex] + 2a(x - x[tex]_{0}[/tex])
V[tex]^{2}[/tex] = (5)[tex]^{2}[/tex] + 2(-9.81)(39.3 - 40)
[tex]\sqrt{V^{2}}[/tex] = [tex]\sqrt{38.688}[/tex]
V = 6.22
Mastering physics says incorrect and to check my signs,, I've checked my signs i can't find my error
edit: ops, since its falling i take the negative root not the positive root, problem solved..answer was -6.27
hello, I've tried this simple problem a few times and mastering physics keeps taking points off =(.
ive never used latex so bare with me
Homework Statement
A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s, releases a sandbag at an instant when the balloon is a height h = 40.0 m above the ground View Figure . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive.
Compute the velocity of the sandbag at a time 1.15 s after its release.
Homework Equations
x = X[tex]_{0}[/tex] +V[tex]_{0}[/tex]T + .5aT[tex]^{2}[/tex]
V[tex]^{2}[/tex] = V[tex]^{2}_{0}[/tex] + 2a(x - x[tex]_{0}[/tex])
The Attempt at a Solution
I assume that the initial velocity of the sandbag once it is released is 5ms since that was the velocity of the balloon.
first i found the position of the sandbag at 1.15.
x = 40 + 5(1.15) + .5(-9.81)(1.15)[tex]^{2}[/tex]
x = 39.3
up to this point i am 100% sure i am correct.
next i inserted this information into the velocity formula.
V[tex]^{2}[/tex] = V[tex]^{2}_{0}[/tex] + 2a(x - x[tex]_{0}[/tex])
V[tex]^{2}[/tex] = (5)[tex]^{2}[/tex] + 2(-9.81)(39.3 - 40)
[tex]\sqrt{V^{2}}[/tex] = [tex]\sqrt{38.688}[/tex]
V = 6.22
Mastering physics says incorrect and to check my signs,, I've checked my signs i can't find my error
edit: ops, since its falling i take the negative root not the positive root, problem solved..answer was -6.27
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