(a) The maximum height reached by the projectile is 10.2 m and
(b) The horizontal distance traveled by the projectile is 40.8 m.
<h3>
Maximum height reached by the projectile</h3>
H = u²sin²θ/2g
H = (20²(sin(45))² / (2 x 9.8)
H = 10.2 m
<h3>Time of motion</h3>
T = (2u sinθ)/g
T = (2 x 20 x sin(45) )/(9.8)
T = 2.886 s
<h3>Horizontal distance traveled by the projectile</h3>
x = (v cos(45) x t
x = (20 x cos(45)) x 2.886
x = 40.8 m
Thus, the maximum height reached by the projectile is 10.2 m and the horizontal distance traveled by the projectile is 40.8 m.
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Answer:
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