Question

Тело падает свободно с высоты 490м. Определите перемещение тела в последнюю секунду падения.

Answer 1

При равноускоренном движении путь: S=a^2*t/2.(a-тэ-квадрат пополам).В данном случае S=490, а=g - ускорение свободного падения , примерно 10 м/сек^2.Тогда время: корень (2*490/10)=9.89 сек.За последнюю секунду путь (из той же формулы):S=S2-S1=g*t^2/2-g*(t-1)^2/2=10/2(9.8^2-8.8^2).Досчитай сам, лень... :-))

Answer 2

Calculating the Displacement of an Object in Free Fall

To determine the displacement of an object in the last second of its fall from a height of 490m, we can use the equations of motion for uniformly accelerated motion.

Given: Initial height (s): 490m Acceleration due to gravity (g): 10 m/s^2

Calculations:

The displacement (S) in the last second of the fall can be calculated using the equation: S = H - (1/2) * g * t^2

Where: - H = Initial height - g = Acceleration due to gravity - t = Time

Using the given values: - H = 490m - g = 10 m/s^2 - t = 1 second (as we are calculating the displacement in the last second of the fall)

Substitute the values into the equation: S = 490 - (1/2) * 10 * 1^2 S = 490 - 5 S = 485m

Answer:

The displacement of the object in the last second of its fall from a height of 490m is 485 meters.