True or false on free fall

Problem

True or false, justifying the answer: (a) a body in free fall has variable acceleration; (b) two bodies dropped from different heights reach the ground with the same velocity; (c) in a rectilinear motion, if the velocity is positive and the acceleration negative, the body is decelerating; (d) if in a $v$–$t$ graph the line has zero slope, the body is at rest.

Solution

(a) FALSE. In free fall the only force is weight, and the acceleration is that of gravity $g \approx 9{,}8\;\mathrm{m/s}^2$, which near the Earth’s surface is constant (in magnitude, direction and sense). What grows during the fall is the velocity, not the acceleration. Free fall is indeed the typical example of uniformly accelerated motion.

(b) FALSE. The velocity with which a body reaches the ground depends on the drop height, according to $v = \sqrt{2 g h}$. A body dropped from a greater height has more room to accelerate and therefore arrives with a greater velocity. Only if the heights were equal would the final velocities coincide.

(c) TRUE. Decelerating means that the magnitude of the velocity decreases over time. This happens when velocity and acceleration have opposite sense: if $v > 0$ and $a < 0$, the acceleration tends to reduce the velocity, which therefore decreases. The body is indeed slowing down.

(d) FALSE. In a $v$–$t$ graph the slope of the line represents the acceleration. Zero slope means zero acceleration, i.e. constant velocity, not necessarily zero velocity. The body can perfectly well move at a constant velocity different from zero: it is at rest only in the particular case where that constant equals zero.

$\ev{\text{(a) F} \quad \text{(b) F} \quad \text{(c) T} \quad \text{(d) F}}$

Collegamenti

Argomenti: Cinematica Concetti: Caduta libera