A hot air balloon is descending at a rate of 2.0 m/s when a passenger drops a camera. (a) If the camera is 40 m above the ground when it is dropped, how longdoes it take for the camera to reach the ground? 1 s (b) What is its velocity just before it lands? Let upward be thepositive direction for this problem. 2 m/s

a.) 1.23 seconds

b.) 14 m/s

Step-by-step explanation:

a.) Before commencing the calculation, we need to specify the information.

dа ta:

acceleration dues to gravity, g = 9.81 m/s²

initial velocity u = 2.0 m/s

height, s = 40 m

t = ?

The formula for finding the distance is s = ut + 1/2at²

Therefore, 40 = 2t + 1/2 * (9.81) * t²

80 = 4t + 9.81 t²

Solving for t by the quadratic equation gives t = 1.23 s [Note the other negative value for t is rejected because there is no negative time]

b) The final velocity is given by the following equation:

v = u + at

where v = final velocity just before the camera lands on the ground

u = initial velocity

t = time taken

a = g = acceleration dues to gravity = 9.81 m/s²

Calculating gives

v = 2 + 9.81*1.23

= 14 m/s Ans