Q.If an object is dropped from the top of a building and it reaches the ground at $t = 4$ s, then the height of the building is (ignoring air resistance, $g = 9.8$ m s$^{-2}$)

Motion Under Gravity – The Intuition First

Drop a ball. It falls. Why? Because the Earth pulls it. But the key insight is not just that it falls — it's how it falls. If you drop a stone from a cliff, does it move at a constant speed? No. It gets faster as it goes down. That's the heart of motion under gravity: the pull is steady, so the acceleration is steady.

Now imagine throwing a ball straight up. It rises, slows down, stops for an instant at the top, then falls back down. The same pull that made it slow down on the way up makes it speed up on the way down. The acceleration never changes direction — it always points downward, toward the centre of the Earth.

That constant downward acceleration is called $g$. Near the Earth's surface, its magnitude is about $9.8\ \text{m/s}^2$ (often taken as $10\ \text{m/s}^2$ in exams for quick calculation). The direction is always downward.

The Precise Statement

Motion under gravity (in one dimension) means an object moves vertically under the sole influence of the Earth's gravitational pull. The acceleration is:

where the negative sign indicates downward direction (if we take upward as positive). The value of $g$ is $9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ in many problems).

Because acceleration is constant, all three equations of uniformly accelerated motion apply directly:

But here $a$ is replaced by $-g$ (if upward is positive) or $+g$ (if downward is positive). The choice of sign convention is yours — just be consistent.

Key Features You Must Know

1. Free Fall (dropped from rest)

If you simply let go of an object ($u = 0$), it falls with increasing speed. After time $t$, its velocity is $v = gt$ (downward). The distance fallen is $s = \frac12 gt^2$.

2. Projected Upward

If you throw a ball upward with speed $u$, it rises until its velocity becomes zero at the highest point. That takes time $t = u/g$. The maximum height reached is $h = u^2/(2g)$.

3. Symmetry of the Motion

The time to go up equals the time to come back down (to the same height). The speed at which it returns to the launch point equals the initial speed $u$ (but downward). This symmetry is a direct consequence of constant acceleration.

A Simple Example …