A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m/s, the (horizontal) distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building? (take $g = 10\ \text{m/s}^2$)

A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof

NCERT Exemplar · Q20

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

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Concept understanding — Projectile Motion Under Gravity

Projectile Motion Under Gravity

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it curves upward, then arcs downward. That curve is a parabola, and the motion is called projectile motion.

The key insight: once the ball leaves your hand, the only force acting on it (ignoring air resistance) is gravity pulling it straight down. There is no force pushing it sideways or upward after release. That single downward force is what creates the beautiful curved path.

The Core Idea

A projectile is any object that is thrown, launched, or otherwise projected into the air and then moves under the influence of gravity alone. The motion has two independent parts happening simultaneously:

Horizontal motion: constant speed (no horizontal force)
Vertical motion: constant downward acceleration $g \approx 9.8 \, \text{m/s}^2$

These two motions are completely independent — they don't affect each other. This is the most important thing to understand.

Important

The horizontal and vertical motions are independent. The horizontal speed stays constant; the vertical speed changes by $9.8 \, \text{m/s}$ every second downward.

Breaking It Down Mathematically

Let's set up coordinates: $x$ is horizontal, $y$ is vertical (positive upward). The launch point is at $(0,0)$ with initial speed $u$ at angle $\theta$ above horizontal.

Initial velocity components:

u_x = u \cos \theta

u_y = u \sin \theta

Horizontal motion (no acceleration):

x = u_x t = (u \cos \theta) t

Vertical motion (constant downward acceleration $g$ ):

y = u_y t - \frac{1}{2} g t^2 = (u \sin \theta) t - \frac{1}{2} g t^2

The minus sign is because gravity pulls downward, opposite to our positive $y$ direction.

The Path Is a Parabola

Eliminate $t$ between the $x$ and $y$ equations. From $x = u_x t$ , we get $t = \frac{x}{u \cos \theta}$ . Substitute into the $y$ equation:

y = (u \sin \theta) \left( \frac{x}{u \cos \theta} \right) - \frac{1}{2} g \left( \frac{x}{u \cos \theta} \right)^2

y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2

This is of the form $y = ax - bx^2$ , which is a parabola opening downward. That's why every projectile under gravity follows a parabolic path.

y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2

Key Quantities You'll Need

Time of Flight ( $T$ )

The total time the projectile stays in the air. Set $y = 0$ (returns to launch height):

0 = (u \sin \theta) T - \frac{1}{2} g T^2

Factor $T$ : $T (u \sin \theta - \frac{1}{2} g T) = 0$

The non-zero solution:

T = \frac{2 u \sin \theta}{g}

Maximum Height ( $H$ )

The highest point occurs when vertical velocity becomes zero: $v_y = u \sin \theta - g t = 0$ , so $t = \frac{u \sin \theta}{g}$ .

Plug into $y$ equation:

H = (u \sin \theta) \left( \frac{u \sin \theta}{g} \right) - \frac{1}{2} g \left( \frac{u \sin \theta}{g} \right)^2

H = \frac{u^2 \sin^2 \theta}{g} - \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 \theta}{2g}

Range ( $R$ )

Horizontal distance traveled when it returns to launch height. Use $x = u_x T$ :

R = (u \cos \theta) \cdot \frac{2 u \sin \theta}{g} = \frac{2 u^2 \sin \theta \cos \theta}{g}

Using $\sin 2\theta = 2 \sin \theta \cos \theta$ :

R = \frac{u^2 \sin 2\theta}{g}

Tip

Maximum range occurs when $\sin 2\theta = 1$ , i.e., $2\theta = 90^\circ$ or $\theta = 45^\circ$ . At this angle, $R_{\text{max}} = \frac{u^2}{g}$ .

Common Mistakes to Avoid

Watch out

Don't mix up horizontal and vertical equations. Horizontal has constant speed; vertical has constant acceleration. …

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The man’s horizontal jump is a projectile under gravity. He clears the 10 m gap only if his time of flight is enough to cover it horizontally. The required time to fall 9 m is about 1.34 s, during which he travels only 12.06 m horizontally — more than 10 m, so he lands safely.

The problem is a classic projectile motion under gravity question, but with a twist: the launch is purely horizontal. That means the initial vertical velocity is zero, and the only force acting is gravity pulling him downward. The horizontal motion, meanwhile, continues at constant speed — no acceleration there.

The key insight: the man’s jump is independent in the vertical and horizontal directions. The time he spends in the air is determined entirely by the vertical drop (9 m). During that time, his horizontal speed (9 m/s) carries him forward. If the horizontal distance covered in that time is at least 10 m, he lands on the next building. If it’s less, he falls short.

Let’s work it through.

Find the time of flight from the vertical drop. The vertical motion starts from rest ( $u_y = 0$ ) and accelerates downward at $g = 10\ \text{m/s}^2$ . The height difference is $h = 9\ \text{m}$ . Using the second equation of motion:

h = \frac{1}{2} g t^2

because initial vertical velocity is zero.

9 = \frac{1}{2} \times 10 \times t^2

9 = 5 t^2

t^2 = \frac{9}{5} = 1.8

t = \sqrt{1.8} \approx 1.3416\ \text{s}

Tip

Notice we didn’t need the mass of the man — gravity acts the same on all objects. This is a pure kinematics problem.

Calculate the horizontal distance covered in that time. Horizontal speed is constant at $v_x = 9\ \text{m/s}$ . So the horizontal range is:

R = v_x \times t = 9 \times \sqrt{1.8}

R \approx 9 \times 1.3416 = 12.0744\ \text{m}

Compare with the gap. …

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Concept: Solving for the Minimum Required Launch Speed, Then Comparing

Method: Isolate the Threshold Speed First (a reusable inequality), Rather Than Computing the Actual Range and Comparing Numbers

The stored answer computes the man's actual horizontal range and checks whether it exceeds the $10\ \text{m}$ gap. This method instead derives the minimum horizontal speed that would just barely clear the gap, as a standalone symbolic threshold, and only then checks the given $9\ \text{m/s}$ against it — a route that immediately generalises to "for a gap $d$ and height difference $h$ , what speed is needed?" without recomputing the fall time and range from scratch each time.

Steps

State the two independent motions. Vertically: free fall from rest, $u_y=0$ , so the time to fall the full height difference $h=9\ \text{m}$ is fixed by $h=\tfrac12gt^2$ , independent of the man's speed:

t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(9)}{10}} = \sqrt{1.8} \approx 1.3416\ \text{s}

Write the condition for "just barely" clearing the gap $d=10\ \text{m}$ . He clears the gap if his horizontal speed $v_x$ , times the time available $t$ , is at least $d$ :

v_x\,t \ge d \quad\Longrightarrow\quad v_x \ge \frac{d}{t}

Solve for the threshold speed directly, substituting $t=\sqrt{2h/g}$ symbolically first (before plugging in numbers):

v_{x,\min} = \frac{d}{t} = \frac{d}{\sqrt{2h/g}} = d\sqrt{\frac{g}{2h}}

This single formula answers "what speed is needed?" for any gap $d$ and height difference $h$ — a reusable threshold, not tied to the specific numbers in this problem.

Substitute the given numbers into the threshold formula:

v_{x,\min} = 10\sqrt{\frac{10}{2(9)}} = 10\sqrt{\frac{10}{18}} = 10\sqrt{0.5556} \approx 10(0.7454) \approx 7.454\ \text{m/s}

Compare the man's actual speed to this threshold — a single inequality, not a distance comparison:

v_x = 9\ \text{m/s} \quad \text{vs.} \quad v_{x,\min} \approx 7.454\ \text{m/s}

Since $9 > 7.454$ , his actual speed exceeds the minimum required speed — he clears the gap. …

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