Ahot-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 (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. a) compute the position of the sandbag at a time 1.05 s after its release. b)compute the velocity of the sandbag at a time 1.05 s after its release. c) how many seconds after its release will the bag strike the ground?

Data:

Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground

Type of motion: free fall.

A) Compute the position of the sandbag at a time 1.05 s after its release.

Equation: y = h + Vo*t - g*(t^2) / 2

y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m

B)Compute the velocity of the sandbag at a time 1.05 s after its release.

Equation: Vf = Vo - g*t

=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s

Negative sign means that the sandbag is going down.

c) How many seconds after its release will the bag strike the ground?

Equation:

y = yo + Vo*t - g*(t^2) / 2

0 = 40.0 + 5.00t - 4.9 t^2

=> 4.9 t^2 - 5t - 40 = 0

Use the quadratic formula and you get: t = 3.41 s