Answer:
minimum muzzle velocity of the dart: [tex]30,97 \frac{m}{s}[/tex]
Explanation:
Al t time t=0, the monkey drops and the dart is shot,
For the monkey the general position equation will be:
[tex]y = y_{0} + V_{0} t + \frac{1}{2} a t^{2}[/tex]
it starts from rest so [tex]V_{0} = 0[/tex], and the acceleration is [tex]a = -g[/tex], so
(1) [tex]y = y_{0} - \frac{1}{2} g t^{2}[/tex]
it will hit the ground when y = 0:
[tex]0 = y_{0} - \frac{1}{2} g t_{ground}^{2}[/tex]
[tex]t_{ground}^{2} = \frac{2*y_{0}}{g}[/tex]
[tex]t_{ground} = 2,26 s[/tex]
For the dart first we need to know the time when it reaches the monkey on the horizontal position, for this we have the equation:
[tex]x = x_{0} + V_{dart} t + \frac{1}{2} a t^{2}[/tex]
In this case, neglecting the air friction acceleration will be 0, and considering the origin on the gun [tex]x_{0} = 0[/tex]:
(2)[tex]x = V_{dart} t[/tex]
So the time when it hits the monkey would be when x = 70m:
(3)[tex]t_{hit} = \frac{70m}{V_{dart}}[/tex]
To hit the monkey before it reaches the ground it must be
[tex]t_{hit} < t_{ground}[/tex]
[tex]\frac{70m}{V_{dart}} < t_{ground}[/tex]
[tex]V_{dart} > \frac{70m}{t_{ground}}[/tex]
[tex]V_{dart} > 30,97 \frac{m}{s}[/tex]
