Question bank — Free fall — g = 9.8 m - s², sign conventions

False (in vacuum). Since $ma=mg$ gives $a=g$, the mass cancels and every object accelerates identically; the "heavier falls faster" effect you see in air is air resistance, not gravity.

False. Free fall means gravity is the only force — this is true on the way up, at the top, and on the way down. The object is decelerating, but only gravity is acting.

False. Only the velocity is momentarily zero; the acceleration is still $-g$. If $a$ were zero the ball would hover there forever instead of coming back down.

False. Gravity points down, so with up positive the acceleration is $a=-g=-9.8{\text{m/s}}^2$. The number $9.8$ is a magnitude; the sign is decided by your chosen axis.

True (no air). By the time-free relation $v^2=v_0^2-2g\Delta y$ with $\Delta y=0$, the return speed magnitude equals $v_0$ — but the velocity is now $-v_0$ (directed downward).

False. It only relabels which direction we call $+$; the actual motion is identical. With down positive we simply write $a=+g$ and downward velocities as positive — see Vectors and sign conventions.

True. Vertical and horizontal motion are independent; the vertical drop obeys $y=y_0-\frac{1}{2}g{t}^2$ regardless of horizontal speed. This is the core idea of Projectile Motion.

False. On the way up the velocity is positive (moving up); it passes through zero at the top and only then becomes negative. The acceleration is negative throughout, not the velocity.

The equation only says velocity is instantaneously zero at that instant $t=v_0/g$; a moment later $t$ is larger so $v$ becomes negative. Zero velocity is a passing instant, not a resting state.

$9.8$ is metres-per-second per second, not per metre. It gains $9.8\text{m/s}$ of speed each second, not each metre of distance dropped.

Acceleration direction and velocity direction are independent. On the way up $a=-g$ while velocity is upward (positive); the negative acceleration is what slows the upward motion.

Signs must match the chosen axis. If up is positive, a downward throw means $v_0=-5$; writing $+5$ mixes conventions and corrupts every later result.

A factor of $\frac{1}{2}$. From $0=v_0^2-2gH$ we get $H=v_0^2/(2g)$; the "2" from integrating the constant acceleration was dropped.

Free fall is defined by the forces, not the direction of motion. Engines off and no air means gravity alone acts, so the coasting rocket IS in free fall even while rising.

Same acceleration, not same impact speed. Impact speed depends on drop height via $v^2=2g\Delta y$; the 20 m ball arrives faster.

Gravity's pull grows with mass ($F=mg$), but so does the resistance to being accelerated ($F=ma$). Setting $ma=mg$ cancels $m$, leaving $a=g$ — the two mass effects exactly offset. See Newton's Second Law.

The motion is symmetric: the same constant acceleration $-g$ that removes speed $v_0$ on the way up restores exactly $v_0$ on the way down over the same height, so both legs take $t=v_0/g$.

Because velocity, displacement, and acceleration are directional (vectors), and a sign is the 1-D way of encoding direction. Without a fixed convention "up" and "down" get mixed and equations give nonsense — Vectors and sign conventions.

Because near Earth's surface $g$ is essentially constant, so $a$ is constant, which is exactly the condition those kinematic equations were derived under.

On Earth air resistance opposes the feather strongly relative to its tiny weight; on the airless Moon that force is gone, so both experience only gravity and fall identically. This is where Air Resistance and Terminal Velocity breaks the ideal.

Acceleration measures how velocity is changing, not its current value. Even at the instant $v=0$, the velocity is on its way from $+$ to $-$, which is a downward (negative) change — hence $a=-g$.

The extra initial speed $v_0$ adds to the final speed through $v^2=v_0^2+2g\Delta y$, but the acceleration stays $g$ because gravity is unchanged — you altered the starting condition, not the force.

Already $g$ downward. Even though speed is zero at $t=0$, gravity acts immediately, so the acceleration is $g$ from the first instant — it never "ramps up".

No. Velocity is zero only at the top, but acceleration is $-g$ there; there is no moment in ideal free fall where the acceleration vanishes.

$v^2=v_0^2$, so $v=\pm v_0$. Physically the meaningful root is $-v_0$ (downward), confirming the ball returns with launch speed but opposite direction.

$v_0=\sqrt{2gH}$ — the same as the speed a dropped ball gains over that height, since the up-trip and a drop over $H$ are mirror images.

Yes, right up to contact the only force is gravity, so $a=g$. The sudden change happens at impact when the ground exerts a normal force, ending the free-fall phase.

It fails: air resistance grows with speed until it balances gravity, giving zero net acceleration and a constant terminal velocity. Free fall assumes this force is absent.

Negative. With down positive, upward motion is the negative direction, so an upward throw begins with $v_0<0$ while $a=+g$; the physics matches up-positive after relabeling.