Acceleration due to Gravity - Worksheet | Science 9th Grade - Free Printable

Problem Analysis and Solution

The worksheet involves problems related to motion under gravity, where air resistance is negligible. We will solve each problem step by step using the equations of motion under constant acceleration due to gravity. The standard equations of motion are:

1. \( v = u + at \)
2. \( s = ut + \frac{1}{2}at^2 \)
3. \( v^2 = u^2 + 2as \)

Where:
- \( u \) = initial velocity
- \( v \) = final velocity
- \( a \) = acceleration (due to gravity, \( g \))
- \( s \) = displacement
- \( t \) = time

On Earth, the acceleration due to gravity is approximately \( g = 9.8 \, \text{m/s}^2 \). On the Moon, we will calculate the acceleration due to gravity based on the given data.

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Problem 1: Determine the speed at impact

#### a) Seagull drops a shellfish from a height of 12.5 m
- Initial velocity, \( u = 0 \, \text{m/s} \)
- Displacement, \( s = 12.5 \, \text{m} \)
- Acceleration, \( a = g = 9.8 \, \text{m/s}^2 \)

We use the equation:
\[ v^2 = u^2 + 2as \]

Substitute the values:
\[ v^2 = 0^2 + 2 \cdot 9.8 \cdot 12.5 \]
\[ v^2 = 2 \cdot 9.8 \cdot 12.5 \]
\[ v^2 = 245 \]
\[ v = \sqrt{245} \]
\[ v \approx 15.65 \, \text{m/s} \]

Answer:
\[ \boxed{15.65 \, \text{m/s}} \]

#### b) Steel ball dropped from the Leaning Tower of Pisa
- Time of fall, \( t = 3.37 \, \text{s} \)
- Acceleration, \( a = g = 9.8 \, \text{m/s}^2 \)

We use the equation:
\[ v = u + at \]

Since the ball is dropped, \( u = 0 \):
\[ v = 0 + 9.8 \cdot 3.37 \]
\[ v = 33.026 \, \text{m/s} \]

Answer:
\[ \boxed{33.03 \, \text{m/s}} \]

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Problem 2: Steel ball thrown from a tower

#### a) Initial velocity upward (\( u = 15.0 \, \text{m/s} \))
- Initial velocity, \( u = 15.0 \, \text{m/s} \) (upward)
- Displacement, \( s = -15.0 \, \text{m} \) (ground level is below the starting point)
- Acceleration, \( a = -g = -9.8 \, \text{m/s}^2 \) (downward)

##### Step 1: Find the total flight time (\( t \))
We use the equation:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ -15.0 = 15.0t + \frac{1}{2}(-9.8)t^2 \]
\[ -15.0 = 15.0t - 4.9t^2 \]
\[ 4.9t^2 - 15.0t - 15.0 = 0 \]

This is a quadratic equation in the form \( at^2 + bt + c = 0 \). Solve using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 4.9 \), \( b = -15.0 \), and \( c = -15.0 \).

\[ t = \frac{-(-15.0) \pm \sqrt{(-15.0)^2 - 4 \cdot 4.9 \cdot (-15.0)}}{2 \cdot 4.9} \]
\[ t = \frac{15.0 \pm \sqrt{225 + 294}}{9.8} \]
\[ t = \frac{15.0 \pm \sqrt{519}}{9.8} \]
\[ t = \frac{15.0 \pm 22.78}{9.8} \]

Take the positive root (since time cannot be negative):
\[ t = \frac{15.0 + 22.78}{9.8} \]
\[ t \approx \frac{37.78}{9.8} \]
\[ t \approx 3.86 \, \text{s} \]

##### Step 2: Find the speed at impact (\( v \))
We use the equation:
\[ v = u + at \]

Substitute the values:
\[ v = 15.0 + (-9.8)(3.86) \]
\[ v = 15.0 - 37.828 \]
\[ v = -22.828 \, \text{m/s} \]

The speed at impact is the magnitude of \( v \):
\[ |v| \approx 22.83 \, \text{m/s} \]

Answers:
- Total flight time: \( \boxed{3.86 \, \text{s}} \)
- Speed at impact: \( \boxed{22.83 \, \text{m/s}} \)

#### b) Initial velocity downward (\( u = -15.0 \, \text{m/s} \))
- Initial velocity, \( u = -15.0 \, \text{m/s} \) (downward)
- Displacement, \( s = -15.0 \, \text{m} \)
- Acceleration, \( a = -g = -9.8 \, \text{m/s}^2 \)

##### Step 1: Find the total flight time (\( t \))
We use the equation:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ -15.0 = -15.0t + \frac{1}{2}(-9.8)t^2 \]
\[ -15.0 = -15.0t - 4.9t^2 \]
\[ 4.9t^2 + 15.0t - 15.0 = 0 \]

Solve using the quadratic formula:
\[ t = \frac{-15.0 \pm \sqrt{(15.0)^2 - 4 \cdot 4.9 \cdot (-15.0)}}{2 \cdot 4.9} \]
\[ t = \frac{-15.0 \pm \sqrt{225 + 294}}{9.8} \]
\[ t = \frac{-15.0 \pm \sqrt{519}}{9.8} \]
\[ t = \frac{-15.0 \pm 22.78}{9.8} \]

Take the positive root:
\[ t = \frac{-15.0 + 22.78}{9.8} \]
\[ t \approx \frac{7.78}{9.8} \]
\[ t \approx 0.794 \, \text{s} \]

##### Step 2: Find the speed at impact (\( v \))
We use the equation:
\[ v = u + at \]

Substitute the values:
\[ v = -15.0 + (-9.8)(0.794) \]
\[ v = -15.0 - 7.7812 \]
\[ v = -22.7812 \, \text{m/s} \]

The speed at impact is the magnitude of \( v \):
\[ |v| \approx 22.78 \, \text{m/s} \]

Answers:
- Total flight time: \( \boxed{0.79 \, \text{s}} \)
- Speed at impact: \( \boxed{22.78 \, \text{m/s}} \)

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Problem 3: Baseball pitcher throws a ball vertically upward

#### a) Initial velocity (\( u \))
- Time of flight, \( t = 4.2 \, \text{s} \)
- The ball returns to the same level, so the total displacement \( s = 0 \).
- Acceleration, \( a = -g = -9.8 \, \text{m/s}^2 \)

At the highest point, the velocity is zero. The time to reach the highest point is half the total time:
\[ t_{\text{up}} = \frac{4.2}{2} = 2.1 \, \text{s} \]

We use the equation:
\[ v = u + at \]

At the highest point, \( v = 0 \):
\[ 0 = u + (-9.8)(2.1) \]
\[ 0 = u - 20.58 \]
\[ u = 20.58 \, \text{m/s} \]

Answer:
\[ \boxed{20.58 \, \text{m/s}} \]

#### b) Maximum height (\( h \))
We use the equation:
\[ v^2 = u^2 + 2as \]

At the highest point, \( v = 0 \):
\[ 0 = (20.58)^2 + 2(-9.8)h \]
\[ 0 = 423.5364 - 19.6h \]
\[ 19.6h = 423.5364 \]
\[ h = \frac{423.5364}{19.6} \]
\[ h \approx 21.61 \, \text{m} \]

Answer:
\[ \boxed{21.61 \, \text{m}} \]

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Problem 4: Hot-air balloon and ballast

#### a) Height of the balloon when the ballast was released
- Initial velocity of the ballast relative to the ground, \( u = 2.1 \, \text{m/s} \) (upward)
- Time of fall, \( t = 3.8 \, \text{s} \)
- Acceleration, \( a = -g = -9.8 \, \text{m/s}^2 \)
- Final displacement, \( s = 0 \) (ballast hits the ground)

We use the equation:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ 0 = 2.1(3.8) + \frac{1}{2}(-9.8)(3.8)^2 \]
\[ 0 = 8.18 - 4.9(14.44) \]
\[ 0 = 8.18 - 70.756 \]
\[ 0 = -62.576 + h \]
\[ h = 62.576 \, \text{m} \]

Answer:
\[ \boxed{62.58 \, \text{m}} \]

#### b) Velocity of the ballast at impact
We use the equation:
\[ v = u + at \]

Substitute the values:
\[ v = 2.1 + (-9.8)(3.8) \]
\[ v = 2.1 - 37.24 \]
\[ v = -35.14 \, \text{m/s} \]

The speed at impact is the magnitude of \( v \):
\[ |v| \approx 35.14 \, \text{m/s} \]

Answer:
\[ \boxed{35.14 \, \text{m/s}} \]

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Problem 5: Astronaut drops a camera on the Moon

#### a) Acceleration due to gravity on the Moon
- Displacement, \( s = 2.3 \, \text{m} \)
- Time of fall, \( t = 1.7 \, \text{s} \)
- Initial velocity, \( u = 0 \, \text{m/s} \)

We use the equation:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ 2.3 = 0(1.7) + \frac{1}{2}a(1.7)^2 \]
\[ 2.3 = \frac{1}{2}a(2.89) \]
\[ 2.3 = 1.445a \]
\[ a = \frac{2.3}{1.445} \]
\[ a \approx 1.6 \, \text{m/s}^2 \]

Answer:
\[ \boxed{1.6 \, \text{m/s}^2} \]

#### b) Ratio of gravitational forces (\( F_{\text{gEarth}} : F_{\text{gMoon}} \))
The gravitational force is given by:
\[ F_g = mg \]

The ratio of gravitational forces is:
\[ \frac{F_{\text{gEarth}}}{F_{\text{gMoon}}} = \frac{mg_{\text{Earth}}}{mg_{\text{Moon}}} = \frac{g_{\text{Earth}}}{g_{\text{Moon}}} \]

Substitute the values:
\[ \frac{F_{\text{gEarth}}}{F_{\text{gMoon}}} = \frac{9.8}{1.6} \]
\[ \frac{F_{\text{gEarth}}}{F_{\text{gMoon}}} \approx 6.125 \]

Answer:
\[ \boxed{6.13} \]

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Final Answers:

1. a) \( \boxed{15.65 \, \text{m/s}} \)
b) \( \boxed{33.03 \, \text{m/s}} \)
2. a) Total flight time: \( \boxed{3.86 \, \text{s}} \), Speed at impact: \( \boxed{22.83 \, \text{m/s}} \)
b) Total flight time: \( \boxed{0.79 \, \text{s}} \), Speed at impact: \( \boxed{22.78 \, \text{m/s}} \)
3. a) \( \boxed{20.58 \, \text{m/s}} \)
b) \( \boxed{21.61 \, \text{m}} \)
4. a) \( \boxed{62.58 \, \text{m}} \)
b) \( \boxed{35.14 \, \text{m/s}} \)
5. a) \( \boxed{1.6 \, \text{m/s}^2} \)
b) \( \boxed{6.13} \)