Answer:
a) V = 46.736m/s
b) θ = 64.66°
c) t ∈ [0, 5.22]s
Explanation:
First we need the instant when the hawk catches the mouse.
For the mouse:
[tex]X_M = V_H*t = 20*t[/tex]
[tex]Y_M = Y_{oM} - \frac{g*t^2}{2}[/tex] [tex]4 = 140 - \frac{10*t^2}{2}[/tex]
From the y-axis equation we get the time of flight of the mouse:
[tex]t = \sqrt{\frac{2*\Delta Y}{g} }=5.22s[/tex]
With this time we calculate the x-position of the mouse:
[tex]X_M = 20*5.22 = 104.4m[/tex]
For the hawk:
[tex]X_H = V_{oH}*to + V_{XDive}*(t-2)=20*2+V_{XDive}*(5.22-2)=X_M[/tex]
Solving for x-component of the dive velocity:
[tex]V_{XDive} = 20m/s[/tex]
On the y-axis:
[tex]Y_H = 140+V_{YDive}*(t-2)=140+V_{YDive}*(5.22-2)=4[/tex]
Solving for the y-component of the dive velocity:
[tex]V_{YDive} = -42.24m/s[/tex]
Now, the speed will be given by (part a):
[tex]V_{Dive}=\sqrt{V_{XDive}^2+V_{YDive}^2}=46.736m/s[/tex]
The angle of the dive will be (part b):
[tex]\theta = atan(\frac{V_{YDive}}{V_{XDive}} )=64.66°[/tex]
For part c, the mouse experienced free fall from t=0 until it was catched at t=5.22s
