Question

Please help my assignment​

Answer 1

Answer:

1)

Final velocity: 25.48 m/s

He fall 33.124 m far.

2)

depth: 57 m

Explanation:

To solve these problems, we can use the equations of motion under constant acceleration due to gravity.

1. For Upton Chuck's free fall:

Given:

• Time, t = 2.60 seconds
• Initial velocity, u = 0 (as he is initially at rest)
• Acceleration due to gravity, g = 9.8 m/s² (assuming near Earth's surface)

We can use the equation of motion:

$\Large\boxed{\boxed{v = u + gt }}$

Where:

• $\bold{\sf v }$ is the final velocity,
• $\bold{\sf u }$ is the initial velocity (which is 0 in this case),
• $\bold{\sf g }$ is the acceleration due to gravity, and
• $\bold{\sf t }$ is the time.

Substituting the values, we get:

$v = 0 + (9.8 \, \textsf{m/s}^2) \times (2.60 \, \textsf{s})$

$v = 25.48 \, \textsf{m/s}$

So, Upton Chuck's final velocity is $\bold{\sf 25.48 \, \textsf{m/s} }$.

To find the distance fallen, we can use the equation:

$\Large\boxed{\boxed{ s = ut + \dfrac{1}{2}gt^2}}$

Where:

• $\bold{\sf s }$ is the distance fallen,
• $\bold{\sf u }$ is the initial velocity,
• $\bold{\sf g }$ is the acceleration due to gravity, and
• $\bold{\sf t }$ is the time.

Since $\bold{\sf u = 0 }$, the equation simplifies to:

$s = \dfrac{1}{2}gt^2$

$s = \dfrac{1}{2} \times (9.8 \, \textsf{m/s}^2) \times (2.60 \, \textsf{s})^2$

$s = 33.124 \, \textsf{m}$

So, Upton Chuck falls approximately $\bold{\sf 33.125 \, \textsf{m} }$.

$\dotfill$

2. For the stone dropped into the well:

Given:

• Time taken to hit the water, $\bold{\sf t = 3.41 }$ s
• Assuming no initial velocity ($\bold{\sf u = 0 }$)
• Acceleration due to gravity, $\bold{\sf g = 9.8 \, \textsf{m/s}^2 }$

We can use the equation for distance fallen:

$\Large\boxed{\boxed{s = ut + \dfrac{1}{2}gt^2 }}$

Substituting the values, we get:

$s = \dfrac{1}{2} \times (9.8 \, \textsf{m/s}^2) \times (3.41 \, \textsf{s})^2$

$s = \dfrac{1}{2} \times 9.8 \times 11.6281 \, \textsf{m}$

$s = 56.97769$

$s = 57 \, \textsf{m ( in nearest whole number)}$

Therefore, the depth of the well is approximately $\bold{\sf 57 \, \textsf{m} }$.