What is the acceleration of an object in free fall near Earth's surface?

Physics

2 answers:

Vitek1552 [10]3 years ago

8 0

Answer:

D 9.8 m/s^2

Explanation:

The force of gravitational gravity on earth is 9.8 m/s^2

Licemer1 [7]3 years ago

8 0

D) 9.8 m/(s^2) is the approximate acceleration of a free falling object near Earth’s surface

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4. A group of students made a rocket and launched it vertically upwards with velocity

Aleks04 [339]

Answer:

Approximately $72.9\; \rm m$ , assuming that the rocket had no propulsion onboard, and that air resistance on the rocket is negligible.

Explanation:

Initial velocity of this rocket: $u = 27\; \rm m\cdot s^{-1}$ .

When the rocket is at its maximum height, the velocity of the rocket would be equal to $0$ . That is: $v = 0\; \rm m \cdot s^{-1}$ .

The acceleration of the rocket (because of gravity) is constantly downwards, with a value of $a = -g = -10\; \rm m \cdot s^{-2}$ .

Let $x$ denote the distance that the rocket travelled from the launch site to the place where it attained maximum height. The following equation would relate $x \!$ to $u$ , $v$ , and $a$ :

$\displaystyle x = \frac{v^2 - u^2}{2\, a}$ .

Apply this equation to find the value of $x$ :

$\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{\left(0\; \rm m\cdot s^{-1}\right)^{2} - \left(27\; \rm m \cdot s^{-1}\right)^{2}}{2 \times 10\; \rm m \cdot s^{-2}} = 36.45\; \rm m\end{aligned}$ .

In other words, the maximum height that this rocket attained would be $36.45\; \rm m$ .

Again, assume that the air resistance on this rocket is negligible. The rocket would return to the ground along the same path, and would cover a total distance of $2\times 36.45\; \rm m = 72.9\; \rm m$ .