A hot-air balloonist, rising vertically with a constant velocity of magnitude V

Question

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 iii above the ground (figure 1) . 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. Part A Compute the position of the sandbag at a time 0.165 s after its release. Part B Compute the velocity of the sandbag at a time 0.165 s after its release. Part C Compute the position of the sandbag at a time 1.35 s after its release.

Explanation / Answer

here we will use the equation of motion to solve this problem :

part 1) we will use :

d = v*t + 1/2 a t^2

values known to us are :

v = 5 m/s
a = - 9.8 m/s^2
t = 0.165 sec

We just need to put in the values in the above eqn :

d = 5* 0.165 + 1/2(-9.8) * 0.165^2 = 0.6925 m ====== 40 m + 0.6925 = 40.69 m ------------->>>>>answer)

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part 2) We need to find the velocity :

values given to us :

vi = 5.00 m/s
t = 0.165 m
a = - 9.81
vf = find

a = (vf - vi)/t
-9.81 = (vf - 5.00) / 0.165
vf = 3.38 m/s ------------------------------------------>>>>>>>>>>>>answer) part b)

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part 3) same as part 1) just need to change the value of t

d = 5* 1.35 + 1/2(-9.8) * 1.35^2 = 2.18025 m =========40 m + 2.18025 m = 42.18025 m ------------>>>>answer) part c )

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