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denis-greek [22]
3 years ago
9

(a) Calculate the height (in m) of a cliff if it takes 2.48 s for a rock to hit the ground when it is thrown straight up from th

e cliff with an initial velocity of 8.19 m/s. m (b) How long (in s) would it take to reach the ground if it is thrown straight down with the same speed
Physics
1 answer:
oee [108]3 years ago
7 0

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;

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

Time to reach the maximum height;

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\\\\

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;

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

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After 2 sec:

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After 3 sec:

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After 5 sec:

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