Let's solve the problem step by step.

Given:

• Height of the table = 82.0 cm = 0.820 m
• Horizontal distance (range) = 96.0 cm = 0.960 m

This is a projectile motion problem where the car is launched horizontally from the table.

(a) Finding the velocity with which the car left the table:

1. Time to fall: Since the car is falling vertically under gravity, we can use the equation: $y = \frac{1}{2} g t^2$ Where:

   • $y = 0.820 \, \text{m}$ (height of the table)
   • $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)
   • $t$ is the time it takes to fall.

   Solving for $t$: $t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 0.820}{9.8}}$

2. Horizontal velocity: Since the horizontal velocity remains constant during the fall, we can use the horizontal distance (range) to calculate it: $v_x = \frac{x}{t}$ Where:

   • $x = 0.960 \, \text{m}$ (horizontal distance)
   • $t$ is the time calculated in step 1.

(b) Angle of velocity just before impact:

The car has two components of velocity just before impact:

• $v_x$ (horizontal velocity) remains the same as the velocity with which it left the table.
• $v_y$ (vertical velocity) can be calculated using: $v_y = g t$

The angle $\theta$ can then be calculated using the formula: $\theta = \tan^{-1} \left( \frac{v_y}{v_x} \right)$

Now, let's calculate these values.### Results: (a) The velocity with which the car left the table is approximately 2.35 m/s.

(b) The angle of the car's velocity with respect to the floor just before impact is approximately 59.66° below the horizontal.

Do you need more details on any of these steps or further clarification?

Here are 5 related questions to expand on this:

1. How would the results change if the height of the table were doubled?
2. What would happen if the horizontal velocity were doubled?
3. How does air resistance affect projectile motion?
4. Can we determine the speed of the car just before it hits the ground?
5. How do you calculate the range of a projectile launched at an angle?

Tip: The angle of velocity upon impact can tell you about the relative magnitudes of horizontal and vertical speeds—if it's closer to vertical, $v_y$ is larger than $v_x$.