Question

An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.12 m/s.
(a) How long after release of the first stone do the two stones hit the water?

(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?

magnitude =

(c) What is the speed of each stone at the instant the two stones hit the water?

first stone =

second stone =

Answer 1

Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

$y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{1}$ - Final height of the first stone, measured in meters.

$y_{1,o}$ - Initial height of the first stone, measured in meters.

$v_{1,o}$ - Initial speed of the first stone, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{1,o} = 47\,m$, $y_{1} = 0\,m$, $v_{1,o} = -2.12\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$, the following second-order polynomial is built:

$-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0$

Roots of the polynomial are, respectively:

$t_{1} \approx 2.866\,s$ and $t_{2}\approx -3.291\,s$

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

$y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}$

$y_{2}$ - Final height of the second stone, measured in meters.

$y_{2,o}$ - Initial height of the second stone, measured in meters.

$v_{2,o}$ - Initial speed of the second stone, measured in meters per second.

$t$ - Time, measured in seconds.

$t_{o}$ - Initial absolute time, measured in seconds.

$g$ - Gravity constant, measured in meters per square second.

Given that $y_{2,o} = 47\,m$, $y_{2} = 0\,m$, $t_{o} = 1\,s$, $t = 2.866\,s$ and $g = -9.807\,\frac{m}{s^{2}}$, the following expression is constructed and the initial speed of the second stone is:

$1.866\cdot v_{2,o}+29.926 = 0$

$v_{2,o} = -16.038\,\frac{m}{s}$

The initial velocity of the second stone is -16.038 meters per second.

c) The final speed of each stone is determined by the following expressions:

First stone

$v_{1} = v_{1,o} + g \cdot t$

Second stone

$v_{2} = v_{2,o} + g\cdot (t-t_{o})$

Where:

$v_{1,o}, v_{1}$ - Initial and final velocities of the first stone, measured in meters per second.

$v_{2,o}, v_{2}$ - Initial and final velocities of the second stone, measured in meters per second.

If $v_{1,o} = -2.12\,\frac{m}{s}$ and $v_{2,o} = -16.038\,\frac{m}{s}$, the final speeds of both stones are:

First stone

$v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)$

$v_{1} = -30.227\,\frac{m}{s}$

Second stone

$v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)$

$v_{2} = -34.338\,\frac{m}{s}$

The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.