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 = 6.00 m/s, releases a sandbag at an instant when the balloon is y = 50.0 m above the ground. 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? m/s What is the greatest height above the ground that the sandbag reaches?

Explanation / Answer

Use the formulas for position and velocity

X = Xo + Vo * t + 0.5 * a * t^2
V = Vo + a * t

X being the final position
V being the final velocity
Xo being the original position
Vo being the original velocity
a being acceleration

a)
X = 50m + (6m/s)(0.25s) + (0.5)(-9.8m/s^2)(0.25s)^2
X = 51.19 m

V = 6m/s + (-9.8 m/s^2)(0.25s)
V = 3.55m/s


X = 50m + (6m/s)(1s) + (0.5)(-9.8m/s^2)(1s)^2
X = 51.1 meters


V = 6m/s + (-9.8m/s^2)(1s)
V = -3.8m/s

b)

You need to make X = 0 and solve for time.
0 = 50 + 6(t) + 0.5*(-9.8)* t^2

solve the quadratic equation

t = 3.86 sec

c)

Using this time we can imput it into the velocity formula.
V = Vo + a t
v = 6 + -9.8 (3.86)
v = -31.82 m/s

d)

To find the greatest height, we need to find the time it takes for the velocity to reach 0 m/s.
So we input 0 into the velocity formula as the final velocity.
0 = 6 + ( -9.8)t
t = 0.612 seconds

plug this into the position formula
x = 50 + 6(0.612) + 0.5 * (-9.8) * (0.612)^2
x = 51.83 meters

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