Free-Fall Acceleration — Question 1

Question 1

Zero velocity at the top is not zero acceleration

A small ball is launched vertically upward from a roof $19.6\ \mathrm{m}$ above level ground with speed $14.7\ \mathrm{m/s}$. Set $t=0$ at release and $y=0$ at ground level, with upward positive. Neglect air resistance and use constant $g=9.8\ \mathrm{m/s^2}$ until first ground contact. The launch and impact interactions are outside this model.

1. Write $y(t)$, $v(t)$ and $a(t)$. Find the time and height of the highest point. Is the ball in free fall while rising and at the top?

2. Solve for ground-contact time, interpret both algebraic roots, and find velocity immediately before contact.

3. Compute displacement and total traveled distance from release to contact. Explain why their magnitudes differ.

4. Rewrite the position and acceleration using a coordinate $z$ that is zero at the roof and positive downward. Check that the impact time is unchanged. Diagnose the statement: “Acceleration reverses sign when the ball starts to fall.”

Question 1 - Solution

Zero velocity at the top is not zero acceleration

(a) Gravity keeps the same direction throughout the flight.

With $t$ in seconds and numerical SI values, $$y=19.6+14.7t-4.9t^2,\quad v=14.7-9.8t,\quad a=-9.8\ \mathrm{m/s^2}.$$ Setting $v=0$ gives $\boxed{t_{\mathrm{top}}=1.50\ \mathrm{s}}$ and $$\boxed{y_{\mathrm{top}}=19.6+\frac{14.7^2}{2(9.8)}=30.625\ \mathrm{m}}.$$ The ball is in free fall on ascent, at the top and on descent. At the top, velocity is instantaneously zero but continues changing at $-9.8\ \mathrm{m/s^2}$; there is no interval of rest.

(b) Enforce the release-to-contact time domain.

Ground contact requires $t^2-3t-4=(t-4)(t+1)=0$. The roots are $4\ \mathrm{s}$ and $-1\ \mathrm{s}$. Only $\boxed{t=4.00\ \mathrm{s}}$ lies after release. The negative root belongs to a backward mathematical extension, not the specified flight. Just before contact, $$\boxed{v=14.7-9.8(4)=-24.5\ \mathrm{m/s}}.$$ Free-fall equations do not determine acceleration during the subsequent impact.

(c) Add lengths on opposite sides of the reversal.

Writing the launch speed as $u=14.7\ \mathrm{m/s}$, the excess distance is twice the upward excursion: $$D-|\Delta y|=41.650-19.600=22.050\ \mathrm{m}=2\frac{u^2}{2g}.$$

The displacement is $\boxed{\Delta y=-19.6\ \mathrm{m}}$. The ascent is $30.625-19.6=11.025\ \mathrm{m}$ and the descent is $30.625\ \mathrm{m}$. Hence $\boxed{D=41.650\ \mathrm{m}}$. Opposite directions cancel in displacement, but both contribute to distance.

(d) A coordinate reversal changes signs, not the event.

Since $z=19.6\ \mathrm{m}-y$, $$z=-14.7t+4.9t^2\quad\text{(metres)},\qquad a_z=+9.8\ \mathrm{m/s^2}.$$ Ground is $z=19.6\ \mathrm{m}$, again reached at 4 s. In a fixed coordinate system acceleration does not reverse at the top. Velocity reverses; the sign of acceleration changes only when the coordinate convention is reversed here.