Aperson holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . when the balloon is 24 m above the ground, she accidentally releases the bag. what is the speed of the bag just before it reaches the ground?

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

  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.