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Sagot :
(a) The plane’s altitude when it released the cargo is 1,960 m.
(b) The speed of the plane when it dropped the cargo is 240 m/s.
The given parameters:
- time of motion of the cargo, t = 20 s
- horizontal distance of the cargo, X = 4800 m
The plane’s altitude when it released the cargo is calculated as follows;
[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0(20) \ + \ \frac{1}{2} (9.8)(20)^2\\\\h = 1,960 \ m[/tex]
The constant horizontal speed of plane when it dropped the cargo is calculated as follows;
[tex]v_x =\frac{X}{t} \\\\v_x = \frac{4800}{20} \\\\v_x =240 \ m/s[/tex]
Learn more about horizontal speed here:https://brainly.com/question/11381799
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