The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ?
NCERT · Chapter 3
Question 3.12
Your route
Think before solving
- 1Identify what is given and what must be found.
- 2Connect the question to the relevant physics idea.
- 3Predict what the answer should depend on.
One helpful hint
Treat 25 m as the maximum allowed height of the projectile, not as its final vertical displacement.
Use the ceiling as a limit
01 · Understand
The ceiling limits maximum height.
02 · Find
Calculate the largest safe angle.
03 · Finish
Use that angle to find range.
Convert the picture into a condition
At the highest point, the projectile must remain at or below the 25 m ceiling.
\[H = \frac{u^2 \sin^2\theta}{2g} \leq 25\,\text{m}\]
Find the maximum safe angle
Use the limiting height to calculate the largest angle that does not hit the ceiling.
\[\sin^2\theta = \frac{2gH}{u^2} \Rightarrow \theta \approx 33.6^\circ\]
Calculate the corresponding range
Use the safe angle in the horizontal-range formula.
\[R = \frac{u^2 \sin 2\theta}{g}\]
Result
Maximum horizontal distance ≈ 150 m
The projectile remains below the 25 m ceiling.
Horizontal range
\[R = \frac{u^2 \sin 2\theta}{g}\]
Variables
- u
- initial velocity
- θ
- angle of projection
- g
- gravitational acceleration
- R
- horizontal range
Use when
Use when launch and landing heights match, air resistance is ignored and acceleration is constant.
Unit check
\[\frac{m^2 s^{-2}}{m s^{-2}} = m\]
Common mistake
Use sin 2θ, not sin²θ.
Maximum height
\[H = \frac{u^2 \sin^2\theta}{2g}\]
Variables
- H
- maximum height
- u
- initial velocity
- θ
- angle of projection
- g
- gravitational acceleration
Use when
Use for a projectile moving under uniform gravity when air resistance is ignored.
Unit check
\[\frac{m^2 s^{-2}}{m s^{-2}} = m\]
Common mistake
Use the vertical component u sin θ, not the complete speed u.
Step-by-Step Solution
Given DataInitial speed of projection, $u = 40\text{ m/s}$Maximum height of ceiling, $H_{\max} = 25\text{ m}$Acceleration due to gravity, $g = 9.8\text{ m/s}^2$Step 1: Determine the Maximum Permissible Launch Angle ($\theta$)The maximum height reached by a projectile is given by the formula:$$H = \frac{u^2 \sin^2\theta}{2g}$$To avoid hitting the ceiling, the maximum height must not exceed $25\text{ m}$:$$25 = \frac{40^2 \sin^2\theta}{2 \times 9.8}$$$$25 = \frac{1600 \sin^2\theta}{19.6}$$$$\sin^2\theta = \frac{25 \times 19.6}{1600} = \frac{490}{1600} = 0.30625$$$$\sin\theta = \sqrt{0.30625} \approx 0.5534$$$$\theta = \arcsin(0.5534) \approx 33.60^\circ$$Step 2: Calculate $\cos\theta$ and $\sin 2\theta$Using trigonometric identities:$$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - 0.30625} = \sqrt{0.69375} \approx 0.8329$$$$\sin 2\theta = 2\sin\theta\cos\theta = 2 \times 0.5534 \times 0.8329 \approx 0.9219$$Step 3: Calculate the Maximum Horizontal Range ($R$)The formula for the horizontal range is:$$R = \frac{u^2 \sin 2\theta}{g}$$Substitute the values:$$R = \frac{40^2 \times 0.9219}{9.8}$$$$R = \frac{1600 \times 0.9219}{9.8} = \frac{1475.04}{9.8} \approx 150.51\text{ m}$$AnswerThe maximum horizontal distance the ball can go without hitting the ceiling is $150.5\text{ m}$ (or $\approx \mathbf{150.7\text{ m}}$ if using $\sin 67.2^\circ \approx 0.923$).
NCERT Class 11 Physics Q3.12 | Why 45° Doesn't Work? | Projectile Motion | Chapter 3 | CBSE
Learn by changing it
Experiment with the Question
Start with this question’s exact values, then change the conditions and see what becomes possible.
Ceiling · 25 m Speed · 40 m s⁻¹ Target · 70 m
Open Projectile Playground →