NDA Physics · Kinematics and Motion

Projectile and Vertical Motion

Vertical throws and projectiles are constant-acceleration motion under gravity; the key idea is that horizontal and vertical motions are independent, so each is handled with the same equations of motion using g.

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Why this matters

Three PYQs, all EASY-to-MODERATE, and they reward a single habit: treat the vertical motion (with acceleration g) and the horizontal motion (constant velocity) separately. Straight-up throws use v² = u² − 2gh and v = u − gt with v = 0 at the top; horizontal projectiles get their fall time from the height alone and their range from that time. Take g = 10 m/s² unless told otherwise, and the marks fall out.

Concept 1 of 2: Vertical throw — straight up under gravity

Throw a ball straight up and gravity decelerates it at g until, at the highest point, its velocity is momentarily zero. The equations of motion apply with a = −g (taking up as positive). The speed it had on the way up is exactly the speed it returns with.

Definition

For a body thrown straight up with speed $u$ (take up as positive, $a = -g$ ):

velocity: $v = u - gt$ , which is 0 at the maximum height;
maximum height: $v^2 = u^2 - 2gh \Rightarrow h_{\max} = \dfrac{u^2}{2g}$ ;
time to the top: $t = \dfrac{u}{g}$ . The ascent and descent times are equal.

Vertical throw (up positive, a = −g)

v = u - gt, \qquad h_{\max} = \dfrac{u^2}{2g}, \qquad t_{\text{up}} = \dfrac{u}{g}

ulaunch speed (upward)
gacceleration due to gravity (≈ 10 m/s²)
$h_{\max}$ maximum height

Worked example

A ball is thrown straight up and reaches a maximum height of 20 m. With what speed was it thrown? (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2021 · Q140Moderate

Example 1 · Kinematics and Motion · Projectile and Vertical Motion

A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of

Velocity is zero at the top, acceleration is NOT

At the highest point the velocity is momentarily 0, but the acceleration is still g downward — gravity never switches off. Use v = 0 only for the velocity, never set a = 0 at the top.

Drill 1 more on vertical throw — straight up under gravity

Concept 2 of 2: Horizontal projectile — independence of motions

Fire a ball horizontally off a cliff and two things happen at once and independently: it keeps a constant horizontal speed (no horizontal force), while gravity pulls it down exactly as if it had been dropped. The fall time depends only on the height; the horizontal range is that time times the launch speed.

Definition

A projectile launched horizontally with speed $u$ from height $h$ :

vertical motion is free fall: time to land $t = \sqrt{\dfrac{2h}{g}}$ (independent of $u$ );
horizontal motion is uniform: range $R = u\,t$ .

The horizontal and vertical motions are independent — they share only the time $t$ .

Horizontal projectile

t = \sqrt{\dfrac{2h}{g}}, \qquad R = u\,t

uhorizontal launch speed
hlaunch height
ttime of flight
Rhorizontal range

Horizontal velocity stays constant; vertical velocity falls to zero at the peak then reverses. The two motions are independent.

Worked example

A stone is thrown horizontally at 12 m/s from the top of a 20 m building. How far from the base does it land? (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2023 · Q52Moderate

Example 2 · Kinematics and Motion · Projectile and Vertical Motion

A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking

g = 10\,\text{m/s}^2

and neglecting air resistance will give :

Horizontal speed never affects the fall time

A faster horizontal throw lands farther away but takes the SAME time to fall, because the vertical motion is independent free fall. Get the time from the height (t = √(2h/g)) first, then multiply by u for the range.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Vertical throw — straight up under gravity
Vertical throw (up positive, a = −g)
$v = u - gt, \qquad h_{\max} = \dfrac{u^2}{2g}, \qquad t_{\text{up}} = \dfrac{u}{g}$
Horizontal projectile — independence of motions
Horizontal projectile
$t = \sqrt{\dfrac{2h}{g}}, \qquad R = u\,t$

Watch out for (2)

Velocity is zero at the top, acceleration is NOT
→ Vertical throw — straight up under gravity
Horizontal speed never affects the fall time
→ Horizontal projectile — independence of motions

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