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 = 4.70 m/s, releases a sandbag at an instant when the balloon is y = 48.0 m above the ground (Fig. 2.37). After it is released, the sandbag is in free fall.

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 4.70 m/s, releases a sandbag at an instant when the balloon is y = 48.0 m above the ground (Fig. 2.37). After it is released, the sandbag is in free fall. Compute the position and velocity of the sandbag 0.250 s after its release Compute the position and velocity 1.00 s after its release How many seconds after its release will the bag strike the ground? With what magnitude of velocity does it strike? What is the greatest height above the ground that the sandbag reaches?

Explanation / Answer

V = at +v0 = speed (positive upwards)

S = 1/2 at^2 +v0.t + s0 (distance above ground)

where a = acceleration due to gravity = -9.81m/s^2 (positive upwards)

v0 = 4.6 m/s (positive upwards)

s0 = height of sandbag (s0 = 39m)

Answers:

(a) velocity + V (at 0.25s) = -9.81 x 0.25 + 4.6 = 2.15m/s
position S (at 0.25s) = -9.81 (0.25)^2/2 + 4.6 x 0.25 + 39 = 39.84m
For the 1.0 second results just replace 0.25 in the above formulae with 1.0 - I guess you can do that!

(b) S = 0 when the bag strikes. i.e -1/2 . 9.81t^2 + 4.6t +39 = 0

i.e. 4.9t^2 - 4.6t - 39 = 0 factorise this quadratic for the solution - it's around 3.3 sec. so I will use this value in what follows:

V at impact = -9.81 . 3.3 + 4.6 = -27.7 m/s

At the greatest height V=0
i.e 9.81t = 4.6 or t =4.6/9.81 = 0.47 sec
plug this in the formula for S and you have

S = -9.81/2 . (0.47)^2 + 4.6 . 0.47 + 39 = 40.08 m

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