Final answer:
The final velocity v of an object dropped from a height of 48.3 meters, using the formula v = √(2gh), is approximately 30.7 m/s when rounded to the nearest tenth.
Step-by-step explanation:
If an object is dropped from a height and it hits the ground with a certain velocity, the relationship between the height h and the final velocity v can be expressed using the physics equation of motion for freely falling objects in Earth's gravitational field. However, there is a typo in the equation provided. The correct equation relating the final velocity v to the height h is v = √(2gh), where g is the acceleration due to gravity (approximately 9.8 m/s2 on the surface of the Earth). To determine the final velocity of an object dropped from a height of h meters, we would substitute h with the given height and solve for v.
Using the corrected equation for an object dropped from a height of 48.3 meters:
- h = 48.3 meters
- g = 9.8 m/s2
- v = √(2 * 9.8 m/s2 * 48.3 m)
- v = √(2 * 9.8 * 48.3)
- v = √(944.28)
- v = 30.7 m/s (to the nearest tenth)
So, when the object hits the ground, it would do so with a velocity of approximately 30.7 m/s, rounded to the nearest tenth.