Answer:
a) 2.7s
b) 29 m/s
Explanation:
The equation for the velocity and position of a free fall are the following
[tex]v=v_{0}-gt[/tex] -(1)
[tex]x=x_{0}+v_{0}t-gt^{2}/2[/tex] - (2)
Since the hot-air ballon is descending at 2.1m/s and the camera is dropped at 42 m above the ground:
[tex]v_{0}=-2.1m/s[/tex]
[tex]x_{0}=42m[/tex]
To calculate the time which it takes to reach the ground we use eq(2) with x=0, and look for the positive solution of t:
[tex]t = \frac{1}{84}(2.1\pm\sqrt{2.1^{2} - 4\times42\times9.81/2} )[/tex]
t = 2.71996
Rounding to two significant figures:
t = 2.7 s
Now we calculate the velocity the camera had just before it lands using eq(1) with t=2.7s
[tex]v=-2.1-9.81*(2.71996)[/tex]
v = -28.782 m/s
Rounding to two significant figures:
v = -29 m/s
where the minus sign indicates the downwards direction
