Refer to the diagram shown below.
In this analysis, wind resistance is ignored, and g = 9.8 m/s².
The meat falls with zero vertical velocity, therefore the time, t, before the meat hits the ground is
[tex] \frac{1}{2}*(9.8 \, \frac{m}{s^{2}})*(t \, s)^{2} = (15 \, m) \\ t= \sqrt{ \frac{15}{4.9} }= 1.75 \, s[/tex]
If the fox catches the meat before it hits the ground, then the fox should travel a horizontal distance d in the same time that the meat travels a horizontal distance (7 -d).
The meat travels a distance of
7 - d = (1.2 m/s)*(1.75 s) = 2.1 m
or
d = 4.9 m
Let v = velocity of the fox when it catches the meat.
If the acceleration of the fox is a m/s², then
v = 1.75a
Also,
[tex]d= \frac{1}{2} *(a \, \frac{m}{s^{2}} )*(1.75 \, s)^{2} = \frac{1}{2}( \frac{v}{1.75})^{2}*(1.75^{2}) \\ 4.9 = 0.875v^{2} \\ v^{2} = 5.6 \\ v = 2.366 \, m/s[/tex]
Answer: 2.37 m/s (nearest hundredth)
