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(a) The time taken be "3.5083 sec".
(b) The distance will be "33.82 m".
(c) The horizontal and vertical components be "9.64 m/s and -22.89 m/s".
According to the question,
Speed, V₀x = 15 Cos (50.0°)
= 9.64 m/s
V₀y = 15 Sin (50.0°)
= 11.49 m/s
Vertical displacement = -20.0 m
(a) We know the relation between displacement, time and acceleration
→ y = vt + [tex]\frac{1}{2}[/tex]at²
By substituting the values,
→ -20.0 = 11.49 t - 4.9 t²
4.9 t² - 11.49 t - 20.0 = 0
t = [tex]\frac{11.49 + \sqrt{132.02+392} }{9.8}[/tex]
= [tex]\frac{11.49 + 22.89}{9.8}[/tex]
= 3.5083 sec
(b) We know the formula,
Distance = Speed × Time
or,
→ X = 9.64 t
= 9.64 × 3.5083
= 33.82 m
(c) Again by using the above relation, we get
Vx = V₀x
= 9.64 m/s
Vy = V₀y - gt
By substituting the values,
= 11.49 - 9.8 × 3.5083
= 11.49 - 34.38
= -22.89 m/s
Thus the response above is correct.
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