Answer :
### Problem 1: Rock Thrown from a Building
A rock is thrown off the top of a 250 m tall building with a horizontal velocity of 8.5 m/s.
#### a. Complete the following table:
[tex]\[ \begin{array}{|l|c|c|} \hline & x\text{-direction} & y\text{-direction} \\ \hline \text{initial velocity} \left( v_1 \right) & 8.5 \, \text{m/s} & 0 \, \text{m/s} \\ \hline \text{acceleration} \left( a \right) & 0 \, \text{m/s}^2 & 9.8 \, \text{m/s}^2 \\ \hline \end{array} \][/tex]
#### b. How long does it take for the rock to reach the ground?
To calculate the time, [tex]\( t \)[/tex], it takes for the rock to reach the ground, we can use the following kinematic equation for the vertical (y) direction:
[tex]\[ y = v_iy \cdot t + 0.5 \cdot a \cdot t^2 \][/tex]
Given:
- Initial vertical velocity, [tex]\( v_iy = 0 \, \text{m/s} \)[/tex]
- Acceleration, [tex]\( a = 9.8 \, \text{m/s}^2 \)[/tex]
- Height, [tex]\( y = 250 \, \text{m} \)[/tex]
The equation simplifies to:
[tex]\[ 250 = 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{250 \times 2}{9.8} \][/tex]
[tex]\[ t^2 = \frac{500}{9.8} \][/tex]
[tex]\[ t^2 \approx 51.02 \][/tex]
[tex]\[ t \approx 7.14 \, \text{s} \][/tex]
#### c. How far away from the base of the building does the rock hit the ground?
To find the horizontal distance, [tex]\( x \)[/tex], the rock travels, we can use the following kinematic equation for the horizontal (x) direction:
[tex]\[ x = v_ix \cdot t \][/tex]
Given:
- Horizontal velocity, [tex]\( v_ix = 8.5 \, \text{m/s} \)[/tex]
- Time, [tex]\( t = 7.14 \, \text{s} \)[/tex]
[tex]\[ x = 8.5 \times 7.14 \][/tex]
[tex]\[ x \approx 60.71 \, \text{m} \][/tex]
### Problem 2: Bullet Shot Horizontally
A bullet is shot horizontally out of a gun that is held 1.25 m above level ground. The bullet hits the ground 1250 m away from where the shooter is standing.
#### a. How long is the bullet in the air?
To calculate the time, [tex]\( t \)[/tex], the bullet is in the air, we can use the following kinematic equation for the vertical (y) direction:
[tex]\[ y = v_iy \cdot t + 0.5 \cdot a \cdot t^2 \][/tex]
Given:
- Initial vertical velocity, [tex]\( v_iy = 0 \, \text{m/s} \)[/tex]
- Acceleration, [tex]\( a = 9.8 \, \text{m/s}^2 \)[/tex]
- Height, [tex]\( y = 1.25 \, \text{m} \)[/tex]
The equation simplifies to:
[tex]\[ 1.25 = 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{1.25 \times 2}{9.8} \][/tex]
[tex]\[ t^2 \approx 0.255 \][/tex]
[tex]\[ t \approx 0.505 \, \text{s} \][/tex]
#### b. What was the initial horizontal velocity of the bullet?
To find the initial horizontal velocity, [tex]\( v_ix \)[/tex], we can use the following kinematic equation for the horizontal (x) direction:
[tex]\[ v_ix = \frac{x}{t} \][/tex]
Given:
- Distance, [tex]\( x = 1250 \, \text{m} \)[/tex]
- Time, [tex]\( t = 0.505 \, \text{s} \)[/tex]
[tex]\[ v_ix = \frac{1250}{0.505} \][/tex]
[tex]\[ v_ix \approx 2474.87 \, \text{m/s} \][/tex]
### Problem 3: Tiger Leaping Horizontally
A tiger leaps horizontally from a 6.5 m high rock with a speed of 3.5 m/s.
#### How far from the base of the rock will she land?
To calculate the time, [tex]\( t \)[/tex], the tiger takes to reach the ground, we can use the following kinematic equation for the vertical (y) direction:
[tex]\[ y = v_iy \cdot t + 0.5 \cdot a \cdot t^2 \][/tex]
Given:
- Initial vertical velocity, [tex]\( v_iy = 0 \, \text{m/s} \)[/tex]
- Acceleration, [tex]\( a = 9.8 \, \text{m/s}^2 \)[/tex]
- Height, [tex]\( y = 6.5 \, \text{m} \)[/tex]
The equation simplifies to:
[tex]\[ 6.5 = 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{6.5 \times 2}{9.8} \][/tex]
[tex]\[ t^2 \approx 1.33 \][/tex]
[tex]\[ t \approx 1.152 \, \text{s} \][/tex]
To find the horizontal distance, [tex]\( x \)[/tex], the tiger covers, we use:
[tex]\[ x = v_ix \cdot t \][/tex]
Given:
- Horizontal velocity, [tex]\( v_ix = 3.5 \, \text{m/s} \)[/tex]
- Time, [tex]\( t = 1.152 \, \text{s} \)[/tex]
[tex]\[ x = 3.5 \times 1.152 \][/tex]
[tex]\[ x \approx 4.03 \, \text{m} \][/tex]
A rock is thrown off the top of a 250 m tall building with a horizontal velocity of 8.5 m/s.
#### a. Complete the following table:
[tex]\[ \begin{array}{|l|c|c|} \hline & x\text{-direction} & y\text{-direction} \\ \hline \text{initial velocity} \left( v_1 \right) & 8.5 \, \text{m/s} & 0 \, \text{m/s} \\ \hline \text{acceleration} \left( a \right) & 0 \, \text{m/s}^2 & 9.8 \, \text{m/s}^2 \\ \hline \end{array} \][/tex]
#### b. How long does it take for the rock to reach the ground?
To calculate the time, [tex]\( t \)[/tex], it takes for the rock to reach the ground, we can use the following kinematic equation for the vertical (y) direction:
[tex]\[ y = v_iy \cdot t + 0.5 \cdot a \cdot t^2 \][/tex]
Given:
- Initial vertical velocity, [tex]\( v_iy = 0 \, \text{m/s} \)[/tex]
- Acceleration, [tex]\( a = 9.8 \, \text{m/s}^2 \)[/tex]
- Height, [tex]\( y = 250 \, \text{m} \)[/tex]
The equation simplifies to:
[tex]\[ 250 = 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{250 \times 2}{9.8} \][/tex]
[tex]\[ t^2 = \frac{500}{9.8} \][/tex]
[tex]\[ t^2 \approx 51.02 \][/tex]
[tex]\[ t \approx 7.14 \, \text{s} \][/tex]
#### c. How far away from the base of the building does the rock hit the ground?
To find the horizontal distance, [tex]\( x \)[/tex], the rock travels, we can use the following kinematic equation for the horizontal (x) direction:
[tex]\[ x = v_ix \cdot t \][/tex]
Given:
- Horizontal velocity, [tex]\( v_ix = 8.5 \, \text{m/s} \)[/tex]
- Time, [tex]\( t = 7.14 \, \text{s} \)[/tex]
[tex]\[ x = 8.5 \times 7.14 \][/tex]
[tex]\[ x \approx 60.71 \, \text{m} \][/tex]
### Problem 2: Bullet Shot Horizontally
A bullet is shot horizontally out of a gun that is held 1.25 m above level ground. The bullet hits the ground 1250 m away from where the shooter is standing.
#### a. How long is the bullet in the air?
To calculate the time, [tex]\( t \)[/tex], the bullet is in the air, we can use the following kinematic equation for the vertical (y) direction:
[tex]\[ y = v_iy \cdot t + 0.5 \cdot a \cdot t^2 \][/tex]
Given:
- Initial vertical velocity, [tex]\( v_iy = 0 \, \text{m/s} \)[/tex]
- Acceleration, [tex]\( a = 9.8 \, \text{m/s}^2 \)[/tex]
- Height, [tex]\( y = 1.25 \, \text{m} \)[/tex]
The equation simplifies to:
[tex]\[ 1.25 = 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{1.25 \times 2}{9.8} \][/tex]
[tex]\[ t^2 \approx 0.255 \][/tex]
[tex]\[ t \approx 0.505 \, \text{s} \][/tex]
#### b. What was the initial horizontal velocity of the bullet?
To find the initial horizontal velocity, [tex]\( v_ix \)[/tex], we can use the following kinematic equation for the horizontal (x) direction:
[tex]\[ v_ix = \frac{x}{t} \][/tex]
Given:
- Distance, [tex]\( x = 1250 \, \text{m} \)[/tex]
- Time, [tex]\( t = 0.505 \, \text{s} \)[/tex]
[tex]\[ v_ix = \frac{1250}{0.505} \][/tex]
[tex]\[ v_ix \approx 2474.87 \, \text{m/s} \][/tex]
### Problem 3: Tiger Leaping Horizontally
A tiger leaps horizontally from a 6.5 m high rock with a speed of 3.5 m/s.
#### How far from the base of the rock will she land?
To calculate the time, [tex]\( t \)[/tex], the tiger takes to reach the ground, we can use the following kinematic equation for the vertical (y) direction:
[tex]\[ y = v_iy \cdot t + 0.5 \cdot a \cdot t^2 \][/tex]
Given:
- Initial vertical velocity, [tex]\( v_iy = 0 \, \text{m/s} \)[/tex]
- Acceleration, [tex]\( a = 9.8 \, \text{m/s}^2 \)[/tex]
- Height, [tex]\( y = 6.5 \, \text{m} \)[/tex]
The equation simplifies to:
[tex]\[ 6.5 = 0.5 \cdot 9.8 \cdot t^2 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{6.5 \times 2}{9.8} \][/tex]
[tex]\[ t^2 \approx 1.33 \][/tex]
[tex]\[ t \approx 1.152 \, \text{s} \][/tex]
To find the horizontal distance, [tex]\( x \)[/tex], the tiger covers, we use:
[tex]\[ x = v_ix \cdot t \][/tex]
Given:
- Horizontal velocity, [tex]\( v_ix = 3.5 \, \text{m/s} \)[/tex]
- Time, [tex]\( t = 1.152 \, \text{s} \)[/tex]
[tex]\[ x = 3.5 \times 1.152 \][/tex]
[tex]\[ x \approx 4.03 \, \text{m} \][/tex]
