Question

What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down?

Short answer

The acceleration of the rock thrown straight upward on the way up will be$\mathbf{-}\mathbf{g}{\mathbf{\text{\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$. At the top, the acceleration will be $\mathbf{g}{\mathbf{\text{\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$. On the way down, the acceleration will be +g m/s².

Step-by-step solution

When the rock is moving upward

There is wind resistance, which operates against the rock in the opposite direction of the airflow (opposite direction of the rock's velocity).

As a result, the rock is being pushed downward by both wind resistance and gravity on the way up.

According to Newton's second law, the combined forces produce a downward acceleration higher than.

When the rock rich top position

The only force acting on the rock at the top of the flight, when its velocity is zero (there is no wind resistance), is gravity. Hence the acceleration will beexactly$\mathbf{g}{\mathbf{\text{\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$ downward.

The rock's velocity is only zero for an infinitesimal amount of time.

When the rock falls down

Gravity still acts downward on the way down, but wind resistance acts upward since the rock's velocity has reversed.

As a result,the combined forces produce a downward acceleration of less than$\mathbf{g}$

You may also use conservation of energy to investigate the wind resistance situation.

Because wind resistance absorbs energy during the rock's flight, you may expect the rock to have less mechanical energy at the end of the flight when it returns to the bat than it had at the beginning.

Because the rock's mechanical energy is kinetic at both the start and finish of its flight, it must travel at a slower speed at the end than it did at the beginning.

That's only conceivable if the average acceleration on the way down is smaller than the average acceleration on the way up.

On the way up, the rock thrown straight upward will accelerate at $\mathbf{-}\mathbf{g}{\mathbf{\text{\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$.The acceleration will be $\mathbf{g}{\mathbf{\text{\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$at the summit. The acceleration will be $\mathbf{+}\mathbf{g}{\mathbf{\text{\hspace{0.17em}m/s}}}^{\mathbf{\text{2}}}$on the way down.