Ask question

Ask question

All categories

3 years ago

11

A man drops a baseball from the top of a building. If the ball is held at a height of 1m before it is dropped, and takes 6.8 sec

onds to hit the ground, how high is the building in meters? (Neglect air resistance)

1 answer:

attashe74 [19]3 years ago

3 0

To solve this problem we will apply the linear motion kinematic equations.

The equation that describes the position as a function of the initial velocity, acceleration and time is given by the relation

$s = v_0 t +\frac{1}{2} at^2$

Here,

$v_0 =$ Initial velocity

t = Time

a = Acceleration, at this case due to gravity

There is not initial velocity then we have that the equation to the given time is

$s = \frac{1}{2} (9.8)(6.8)^2$

$s = 226.8m$

If the ball is held at a height of 1m before it is dropped, we have that the Building height is

$h = 226.8-1$

$h = 225.8m$

You might be interested in

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago

Plate Tectonics Lab Report The outcome variable (dependent variable): the outcome/dependant variable is Test variable (independe

ANTONII [103]

Answer:Test variable (independent variable): the land of the 6 locations

Outcome variable (dependent variable): the location

5 0

2 years ago

2) A constant net force acts on an object. Describe the motion of the object.?

77julia77 [94]

Constant acceleration

7 0

3 years ago

A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi

Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

$k = \frac{155}{0.10} N/m$

$k = 1550 N/m$

so now we have

$2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}$

$m = 8.11 kg$

so mass of the fish is 8.11 kg

4 0

3 years ago

How does gay-lussac's law apply to everyday life?

vladimir2022 [97]

Gay-Lussac's Law shows the direct relationship between pressure and temperature for an ideal gas with constant volume. Mathematically it is
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

This particular has a lot of application in our everyday life. In cooking, for example, we apply this concept when using a pressure cooker. We increase/decrease the temperature to meet the right amount of pressure.

In addition, knowing that pressure increases when temperature does can help you with road safety. Knowing that temperature affects heat directly, we must be careful in making sure that tires are not overheated or else they explode out of too much pressure inside.

5 0

3 years ago