Help with Physics?

Question

A hot air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s releases a sandbag at an instant when the balloon is 40.0m above the ground. After it is released, the sandbag is in freefall.
a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release.
b) How many seconds after its release will the bag strike the ground?
c) With what magnitude of velocity does it strike?

Answer 1

Well, we know the following information

initial height = 40 m
initial velocity = 5 m/s
acceleration = -9.8 m/s/s

The kinematic equations will come to our aid for this one.

hf = 1/2 g * t^2 + vi * t + hi

where
hf = final height
g = acceleration due to gravity
t = time
vi = initial velocity
hi = inital height

a) t = 0.250 s

hf = 1/2 g * t^2 + vi * t + hi
hf = 1/2 * -9.8 * (0.250)^2 + 5 * 0.250 + 40
hf = 40.9 m

t = 1.00 s
hf = 40.1m

b) find t when hf = 0
hf = 1/2 g * t^2 + vi * t + hi
0 = 1/2 * -9.8 * t^2 + 5 * t + 40
0 = -4.9t^2 + 5t + 40

Using the quadratic formula,
t = 3.413 s (3.41 s with sig figs)

c) Use the formula

vf ^2 = vi^2 + 2ax

vf ^ 2 = 5^2 + 2 * -9.8 * -40
vf ^ 2 = 809
vf = 28.44 m/s downward

So, the bag was travelling 28.4 m/s down at the moment of impact.

Notice I used an equation which did not require time to answer the question. The time to hit the ground was a calculated value. If we were wrong, then every answer using it would be wrong as well.

As a check, we can use our answers in the following equation.

vf = vi + a*t
-28.42 = 5 + -9.8 * 3.41
-28.42 = -28.42

Check!