Chapter 3, Problem 12TE

What is the acceleration of a ball at the top of its trajectory when thrown straight upward? Explain whether or not your answer is zero by the using the equation a=F/m as a guide to your thinking. (Interesting, almost everyone gets the wrong answer unless they choose F to be mg and let Newton's second law guide their thinking! Use this question if you're helping someone make the distinction between speed and acceleration.)

To determine

The acceleration of a ball at the top of its trajectory when thrown straight upward.

Answer to Problem 12TE

The acceleration of a ball at the top of its trajectory when thrown straight upward is $-9.81{\text{m/s}}^2$.

Explanation of Solution

The acceleration is the variation of velocity concerning time. The magnitude of the acceleration is zero if the velocity remains constant. It depends on the direction of the motion. Acceleration can be determined by the relation of the applied force on the body and it’s mass.

The gravitational pull of the planet only influences the acceleration of the free fall body, and its magnitude is equal to the acceleration due to gravity. The ball is thrown straight upward, and at the top of its trajectory its velocity becomes zero, and the net acceleration is $-9.81{\text{m/s}}^2$. After that, when it falls back, the earth exerts $-9.81{\text{m/s}}^2$ acceleration over it. So, the acceleration of a ball at the top of its trajectory when thrown straight upward is $-9.81{\text{m/s}}^2$.

Conclusion:

Therefore, the acceleration of a ball at the top of its trajectory when thrown straight upward is $-9.81{\text{m/s}}^2$.