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3 years ago

5

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

1 answer:

ziro4ka [17]3 years ago

3 0

$d = u \times t \: + \frac{1}{2} \times a \times {t}^{2}$
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2

$d = 0.5 \times 9.8 \times {4.2}^{2}$
on solving we get

d = 86.436 metres

Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2

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Answer

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A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward

algol [13]

Answer:

a) $v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s$

b) $v=+9.97m/s$

Explanation:

From the exercise we know that

$x_{1} =15m, t_{1}=3s$

$x_{2} =-3m, t_{1}=1.74s$

$x_{3} =29m, t_{3}=5.20s$

From dynamics we know that the formula for average velocity is:

$v=\frac{x_{2}-x_{1} }{t_{2}-x_{1} }$

a) For the three intervals:

$v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s$

$v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s$

$v_{3}=\frac{x_{3}-x_{1} }{t_{3}-t_{1} }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s$

b) The average velocity for the entire motion can be calculate by the following formula:

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3 years ago