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

Intuition

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)

At the top $v = 0$ ; use $v^2 = u^2 - 2gh$ ⟹ $0 = u^2 - 2gh$ .
$u = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400}$ .
$u = 20$ m/s.

Answer:20 m/s.

Practice this conceptself-check · 4 quick reps

Try it yourself

A ball is thrown straight up at 40 m/s. How long does it take to reach the highest point? (g = 10 m/s²)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Thrown up at 30 m/s. Max height? (g = 10)
2.
Thrown up at 20 m/s. Time to top? (g = 10)
3.
At the highest point of a vertical throw, the velocity is?
4.
Reaches 5 m max height. Launch speed? (g = 10)

From the bank · past-year question

Example 1Kinematics and MotionMODERATE

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

[Q140 · Sep · 2021]

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

Intuition

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)

Fall time from the height alone: $t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2$ s.
Horizontal range: $R = u\,t = 12 \times 2 = 24$ m.

Answer:24 m.

Practice this conceptself-check · 4 quick reps

Try it yourself

A ball is rolled off a 5 m table at 4 m/s. How far from the table does it land? (g = 10 m/s²)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Thrown horizontally from 45 m. Time to land? (g = 10)
2.
Two balls — one dropped, one thrown horizontally — leave the same height together. Which lands first?
3.
Range if u = 10 m/s and flight time 2 s?
4.
Does the fall time of a horizontal projectile depend on its launch speed?

From the bank · past-year question

Example 2Kinematics and MotionMODERATE

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 :

[Q52 · Apr · 2023]

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

Mastery check — 1 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Kinematics and MotionEASY

A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately

[Q69 · Apr · 2022]

Drill every past-year question on this subtopic

3 questions from the bank — paginated, with cart and Word-export support.

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