A car falls off a ledge and drops to the ground in 0.6 s. Take (g = 10 , $$\text{m/s}^2$$​). What is the car’s speed on striking the ground?

Correct option is D

The correct answer is (D) 6 m/s

Explanation:
• This problem can be easily solved by applying the fundamental equations of motion for an object moving under a uniform acceleration, which in this case is the acceleration due to gravity (g).
• When the car falls off the ledge, it starts from a state of rest. Therefore, its initial velocity (u) is equal to $$0\text{ m/s}$$​. The time taken (t) to hit the ground is given as $$0.6\text{ seconds}$$​, and the downward acceleration due to gravity (g) is specified as $$10\text{ m/s}^2$$​.
• To find the final speed (v) of the car as it strikes the ground, we use the first equation of motion: v = u + at. Since the car is falling freely under gravity, we replace the general acceleration 'a' with 'g', modifying the formula to v = u + gt.
• Substituting the given values into the equation yields: $$v = 0 + (10 \times 0.6)$$​. Solving this simple multiplication gives $$v = 6\text{ m/s}$$​. Thus, the speed of the car at the exact instant it touches the ground is exactly $$6\text{ m/s}$$​.

Information Booster:
• In physics, whenever an object undergoes free fall near the surface of the Earth, its speed increases uniformly by approximately $$9.8\text{ m/s} (or 10\text{ m/s}$$​ when rounded for computational ease) during every elapsed second of its descent.
• We can also calculate the height of the ledge using the second equation of motion, $$s = ut + \frac{1}{2}gt^2$$​. Substituting the values gives $$s = 0 + \frac{1}{2} \times 10 \times (0.6)^2 = 5 \times 0.36 = 1.8\text{ meters}$$​. This comprehensive analysis allows us to understand the entire physical scenario of the drop.

Additional Knowledge:
• 10 m/s (Option A): This speed would be attained by the falling car if it were allowed to fall freely for a full $$1.0\text{ second} (v = 0 + 10 \times 1 = 10\text{ m/s})$$​.
• 3 m/s (Option B): This represents the intermediate speed of the car exactly halfway through its fall, corresponding to a time of $$$0.3\text{ seconds} (v = 0 + 10 \times 0.3 = 3\text{ m/s})$$​.
• 12 m/s (Option C): This speed would be reached by the car if the fall lasted twice as long, specifically for a duration of $$1.2\text{ seconds} (v = 0 + 10 \times 1.2 = 12\text{ m/s}$$​).