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Answer:
(a) Time flight is 6.39 s
(b) The range of the projectile is 639 m.
Explanation:
Given;
initial horizontal velocity of the projectile, [tex]v_x[/tex] = 100 m/s
initial vertical velocity of the projectile, [tex]v_y[/tex] = 0
height of the tower, h = 200 m
(a) Time flight is calculated as follows;
[tex]h = v_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 200}{9.8} }\\\\t = 6.39 \ s[/tex]
(b) The range of the projectile is calculated as follows;
[tex]R = v_x \times t\\\\R = 100\ m/s \ \times \ 6.39\ s\\\\R = 639 \ m[/tex]