To solve this problem, we will analyze the motion of the block separately in the horizontal and vertical directions.
Given:
- Height of the tabletop (h) = 1.2 m
- Initial horizontal velocity (u_x) = 4 m/s
- Acceleration due to gravity (g) = 9.81 m/s²
Part (a): Horizontal Distance from the Table
First, we need to determine how long it takes for the block to fall to the ground (time of flight). The time of flight can be calculated using the formula for vertical motion under gravity:
\[ h = \frac{1}{2} g t^2 \]
Rearranging for \( t \):
\[ t^2 = \frac{2h}{g} \]
Substituting the values:
\[ t^2 = \frac{2 \times 1.2 \text{ m}}{9.81 \text{ m/s}^2} \approx \frac{2.4}{9.81} \approx 0.2447 \] \[ t = \sqrt{0.2447} \approx 0.4947 \text{ s} \]
Now that we have the time of flight, we can calculate the horizontal distance traveled during this time:
\[ \text{Horizontal distance} = u_x \cdot t \]
Substituting the known values:
\[ \text{Horizontal distance} = 4 \text{ m/s} \cdot 0.4947 \text{ s} \approx 1.9788 \text{ m} \]
So, the horizontal distance from the table at which the block strikes the floor is approximately 1.98 m.
Part (b): Horizontal and Vertical Components of Velocity When It Reaches the Floor
- Horizontal Component of Velocity (\(v_x\)): The horizontal component of velocity remains constant (since there is no horizontal acceleration in this scenario):
\[ v_x = u_x = 4 \text{ m/s} \]
- Vertical Component of Velocity (\(v_y\)): The vertical component of velocity just before it hits the ground can be calculated using the formula for vertical motion:
\[ v_y = g \cdot t \]
Substituting the values:
\[ v_y = 9.81 \text{ m/s}^2 \cdot 0.4947 \text{ s} \approx 4.86 \text{ m/s} \]
Summary of Results:
- Horizontal distance from the table: \(\approx 1.98 \text{ m}\)
- Horizontal component of velocity when it reaches the floor: \(4 \text{ m/s}\)
- Vertical component of velocity when it reaches the floor: \(\approx 4.86 \text{ m/s}\)
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