Let's solve the problem step-by-step.

(a) How long does it take for the car to fall?

To find the time it takes for the car to fall, we can use the vertical motion equation under gravity:

$y = \frac{1}{2} g t^2$

where:

• $y$ is the vertical distance fallen (1.352 m),
• $g$ is the acceleration due to gravity (9.8 m/s²),
• $t$ is the time in seconds.

Rearranging for $t$, we get:

$t = \sqrt{\frac{2y}{g}}$

Substituting the values:

$t = \sqrt{\frac{2 \times 1.352}{9.8}}$

Calculating:

$t \approx 0.525 \, \text{s}$

So, it takes approximately 0.525 seconds for the car to fall.

(b) What is the horizontal velocity of the car?

The horizontal velocity can be found by using the formula:

$v_x = \frac{x}{t}$

where:

• $x$ is the horizontal distance traveled (0.426 m),
• $t$ is the time it took to fall (0.525 s).

Substituting the values:

$v_x = \frac{0.426}{0.525}$

Calculating:

$v_x \approx 0.811 \, \text{m/s}$

So, the horizontal velocity of the car is approximately 0.811 m/s.

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Summary of Answers:

• (a) Time to fall: $\approx 0.525$ seconds
• (b) Horizontal velocity: $\approx 0.811 \, \text{m/s}$

Would you like further details or explanations?

Related Questions:

1. How would the time to fall change if the table height was doubled?
2. What would be the horizontal velocity if the car landed 1 m from the base of the table?
3. How does air resistance affect the car's trajectory?
4. How would the time change if gravity was weaker, say on the Moon?
5. What is the final vertical velocity of the car just before it hits the ground?

Tip:

In projectile motion, horizontal and vertical motions are independent, allowing us to calculate each separately based on initial conditions.