Physics, 03.11.2020 16:20, mimithurmond03

A stone is thrown from the top of a building with an initial velocity of 22.0m/s straight upward, at an initial height of 45.5m above the ground. The stone just misses the edge of the roof on its way down. Determine (a) the time needed for the stone to reach its maximum height, (b) the maximum height, (c) the time needed for the stone to return to the height from which it was originally thrown, and (d) the time needed for the stone to reach the ground.

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answered: Sambo2239

a) The stone will need 2.243 seconds to reach its maximum height.

b) The maximum height reached by the stone is 70.176 meters.

c) The stone will return to the height from which it was originally thrown in 4.486 seconds.

d) The stone needs approximately 6.026 seconds to reach the ground.

Explanation:

a) According to the statement, the stone experiments a free fall, that is, an uniform accelerated motion due to gravitation and in which effects from Earth's rotation and air friction are neglected. In this case, we need to find the time needed for the stone to reach its maximum height by using the following kinematic equations:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g\cdot t^{2}$ (Eq. 1)

$v = v_{o}+g\cdot t$ (Eq. 2)

Where:

$y_{o}$ , - Initial and final heights, measured in meters.

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

- Time, measured in seconds.

- Gravitational constant, measured in meters per square second.

The stone reaches the maximum height when its velocity is zero. If we know that $v_{o} = 22\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then we calculate the time needed for the stone to reach its maximum height:

$t = \frac{v-v_{o}}{g}$

$t = \frac{0\,\frac{m}{s}-22\,\frac{m}{s} }{-9.807\,\frac{m}{s^{2}} }$

$t = 2.243\,s$

The stone will need 2.243 seconds to reach its maximum height.

b) From (Eq. 1) and given that $y_{o} = 45.5\,m$ , $v_{o} = 22\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t = 2.243\,s$ , we find that maximum height of the stone is:

$y = 45.5\,m + \left(22\,\frac{m}{s} \right)\cdot (2.243\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.243\,s)^{2}$

$y = 70.176\,m$

The maximum height reached by the stone is 70.176 meters.

c) The time needed for the stone to return to the height from which it was originally thrown is obtained by solving (Eq. 1) under the following conditions: ( $y = 45.5\,m$ , $y_{o} = 45.5\,m$ , $v_{o} = 22\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ )

$22\cdot t -4.904\cdot t^{2} = 0$ (Eq. 3)

$4.904\cdot t\cdot (4.486-t) = 0$

There are two solutions: $t_{1} = 0\,s$ , $t_{2} = 4.486\,s$ . The first one represents the moment when the stone is thrown, whereas the second one is the instant when stone returns to the hight from which it was originally thrown. Hence, the stone will return to the height from which it was originally thrown in 4.486 seconds.

d) The time needed for the stone to reach the ground is obtained by solving on (Eq. 1) under the following conditions: ( $y = 0\,m$ , $y_{o} = 45.5\,m$ , $v_{o} = 22\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ )

$-4.904\cdot t^{2}+22\cdot t +45.5 = 0$ (Eq. 4)

Then, we solve the resulting expression by Quadratic Formula and find respective roots:

$t_{1}\approx 6.026\,s$ , $t_{2}\approx -1.540\,s$

Only the first roots offers a solution that is physically reasonable. Therefore, the stone needs approximately 6.026 seconds to reach the ground.

answered: Guest

i guess that visual aids present persuasive examples to what you are presenting, some factors would be theme (does it fit with the color, information, what you are presenting), size (can you see it, is it a big enough span of data/information), and overall look of the visual aid.

hope this : )

answered: Guest

the rearrangement of particles in a physical change. hope this

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A stone is thrown from the top of a building with an initial velocity of 22.0m/s straight upward, at...