Question

What is the acceleration of a rock at the top of its trajectory when thrown straight upward. explain whether or not the answer is zero by using the equation a = f/m as a guide to your thinking?

Answer 1

Answer:

Acceleration $a = 9.8 m/s^2$ directed downward

Explanation:

From Newton's second law we know that

$F = ma$

so

$a = \frac{F}{m}$

The force acting on the abject is the gravitational pull of the earth. Which is also called the weight of the object

$a = \frac{F}{m} \\\\a = \frac{W}{m}$

We know that

$W = mg$

So

$a = \frac{mg}{m} \\\\a = g\\\\a = 9.8 m/s^2$

Answer 2

The acceleration of the rock is the gravitational acceleration due to earth which is around 9.8 m/s^2. As the rock travels upwards it is working against the constant force of gravity and when it reaches the top it is the force of gravity that will cause a cause in the direction of the rocks velocity and cause it to accelerate towards earth. Using the equation a = F/m, the acceleration due to gravity exerts a force per unit mass on an object causing the object to accelerate, anything above earth surface will therefore have an external gravitational force action on it (in terms of projectile motion). So at the top of the rocks trajectory, the acceleration of the rock will be g (9.8) as a constant force is gravity is exerted on the rock throughout its path.