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Answer:
(a) The height of the cliff is 9.92 m
(b) Time taken to reach the ground is 0.81 s
Explanation:
Given;
time of motion of the rock, t = 2.48 s
initial velocity of the rock, u = 8.19 m/s
(a) This is a two-way problem;
maximum upward height reached by the rock (h) + Total distance traveled downwards from the top (H).
Height of cliff = H - h
Maximum height reached by the rock;
[tex]v^2 = u^2 - 2gh\\\\0 = (8.19)^2 - (2\times 9.8)h\\\\0 = 67.08 - 19.6 h\\\\19.6h = 67.08\\\\h = \frac{67.08}{19.6} \\\\h = 3.422 \ m[/tex]
Time to reach the maximum height;
[tex]h = ut - \frac{1}{2} gt^2\\\\3.422 = 8.19t - (0.5 \times 9.8)t^2\\\\3.422 = 8.19t - 4.9t^2\\\\4.9t^2 -8.19t + 3.422 = 0\\\\This \ forms \ a quadratic \ equation ; \\\\a = 4.9, b = -8.19, c = 3.422\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 -4ac} }{2a} \\\\t = \frac{-(-8.19) \ \ +/- \ \ \sqrt{(-8.19)^2 -4(4.9\times 3.422)} }{2(4.9)} \\\\t = \frac{8.19 \ \ + /- \ \ \sqrt{0.0049} }{9.8} \\\\t = \frac{8.19 - 0.07}{9.8} \ \ or \ \ t = \frac{8.19 + 0.07}{9.8} \\\\t = 0.83 \ s \ \ or \ \ t = 0.84 \ s\\\\[/tex]
The time taken for rock to falll to the gound;
t₂ = 2.48 - 0.83 s
t₂ = 1.65 s
The distance traveled downwards is calculated as
H = vt + ¹/₂gt²
where;
v is the final velocity at maximum height = initial velocity when the rock starts moving downwards = 0
H = ¹/₂gt²
H = 0.5 x 9.8 x (1.65)²
H = 13.34 m
The height of the cliff = 13.34 m - 3.422m = 9.92 m
(b) time taken to reach the ground if thrown with the same speed;
h = ut + ¹/₂gt²
9.92 = 8.19t + (0.5 x 9.8)t²
9.92 = 8.19t + 4.9t²
4.9t² + 8.19t - 9.92 = 0
a = 4.9, b = 8.19, c = -9.92
use the quadratic formula;
[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 -4ac} }{2a} \\\\t = \frac{-8.19 \ \ +/- \ \ \sqrt{(8.19)^2 -4(4.9\times -9.92)} }{2(4.9)} \\\\t = 0.81 \ s[/tex]