A balloon is ascending at 12.4m/s at a height of 81.3m above the ground when a package is dropped. a) How long did it take to reach the ground

3secs

Explanation:

Given the following parameters

height H= 81.3m

Velocity v = 12.4m/s

Required

Time it take to reach the ground

Using the equation of motion

H = ut+1/2gt²

81.3 = 12.4t + 1/2(9.8)t²

81.3 = 12.4t + 4.9t²

4.9t² + 12.4t - 81.3 = 0

Using the general formula to find t

t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)

t = -12.4±√153.76+1593.48/2(4.9)

t = -12.4±√1747.24/9.8

t = -12.4+41.8/9.8

t = 29.4/9.8

t = 3secs

Hence it took 3secs to reach the ground

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.