A balloon is launched straight upward from the edge of a 40 m high cliff with a velocity of 30 m/s. The balloon goes up and barely misses the edge of the cliff on the way down and hits the
ground below.
A.) Determine the overall hang time of the balloon.
B.) How high did the balloon travel above the cliff?
C.) Determine the velocity of the balloon at t = 3 s.
D.) Determine the acceleration of the balloon at t = 3 s.

A)  t = 7.25 sec

B) 45.92 m above the cliff and 85.92 m counting from ground level

C) about 0.6 m/s

D) The only acceleration is always that of gravity : 9.8 m/s^2

Explanation:

The kinematic equation for the position as a function of time is:

y = 40 + 30 * t - (g/2) t^2

The total time the balloon was in the air is calculated for when the final position is y = 0 (when it touches the ground), that is:

0 = 40 + 30 * t - (g/2) t^2

0 = 40 +30 t - 4.9 t^2

which can be solved for "t" using the quadratic formula, and which renders two different times "t", of which we pick the positive t = 7.25 sec.

For the maximum height, we estimate the time it takes for the balloon to reach the maximum height at which the velocity is zero (changes direction of motion):

vf = vi - g * t

0 = 30 - 9.8 * t

t = 30 / 9.8 = 3.06 seconds

Now we use this value in our position equation and get:

ymax =  40 + 30 (3.06) - 4.9 (3.06)^2

ymax = 85.92 m  (from ground level

Therefore, from the cliff (that is at 40 m height) the height is 85.92 - 40 = 45.92 m.

The velocity of the balloon at t = 3 seconds can be calculated with the velocity expression we have been using:

v(t) = vi - g * t

v(3) = 30 - 9.8 (3) = 0.6  m/s

The only acceleration acting on the balloon is always the acceleration due to gravity (g = 9.8 m/s^2)

need to see circuit to get time constant