Explanation:
The final velocity of the bale have vertical and horizontal components. So, we have:
[tex]v^2=v_x^2+v_y^2(1)[/tex]
According to conservation of energy:
[tex]\frac{mv_y^2}{2}=mgh\\v_y=\sqrt{2gh}[/tex]
Since the plane is flying due east, we have [tex]v_x=49\frac{m}{s}[/tex]. Replacing in (1):
[tex]v=\sqrt{(\sqrt{2gh})^2+v_x^2}\\v=\sqrt{2(9,81\frac{m}{s^2})(66m)+(49\frac{m}{s})^2}\\v=60.78\frac{m}{s}[/tex]
Now, we calculate the momentum of the bale:
[tex]p=mv\\p=\frac{160N}{9.81\frac{m}{s^2}}(60.78\frac{m}{s})\\p=991.32\frac{kg\cdot m}{s}[/tex]
From pythagoras theorem we know that:
[tex]tan\theta=\frac{opposite}{adjacent}\\tan\theta=\frac{v_x}{v_y}\\\theta=tan^{-1}(\frac{v_x}{v_y})\\\theta=tan^{-1}(\frac{60,78\frac{m}{s}}{49\frac{m}{s}})\\\theta=51.12^\circ[/tex]
(a) The momentum of the bale the moment it strikes the ground is 587.4 Kgm/s.
(b) The angle the bale will strike the ground is 36.3⁰.
The given parameters;
The mass of the bale is calculated as;
[tex]W = mg\\\\m = \frac{W}{g} \\\\m = \frac{160}{9.8} \\\\m = 16.33 \ kg[/tex]
The time of motion of the bale is calculated as;
[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 66 }{9.8} } \\\\t = 3.67 \ s[/tex]
The final velocity of the bale is calculated as;
[tex]v_f = v_o_y + gt\\\\v_f = 0 + 3.67 \times 9.8\\\\v_f = 35.97 \ m/s[/tex]
The momentum of the bale the moment it strikes the ground is calculated as;
[tex]P = mv\\\\P = 16.33 \times 35.97\\\\P = 587.4 \ kgm/s[/tex]
The angle the bale will strike the ground is calculated as;
[tex]tan(\theta ) = \frac{v_f_y }{v_f_x} \\\\\theta = tan^{-1} (\frac{v_f_y }{v_f_x} )\\\\\theta = tan^{-1} (\frac{35.97 }{49} )\\\\\theta = 36.3 \ ^0[/tex]
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