Class 9 Science | Gravitation | Ex Q13 | Ball Thrown Vertically | NCERT Step-by-Step Solution
🌍 Class 9 | Science | Chapter 9 – Gravitation
📘 NCERT Exercise | Question 13 | Step-by-Step Solution | CBSE 2026
In this video, Shivang Aggarwal Sir (Saraswati Vidyamandir) explains *Question 13* from *NCERT Class 9 Science Chapter – Gravitation*.
This question deals with the *motion of a ball thrown vertically upwards* and involves the use of equations of motion under gravity.
🧮 Question:
A ball is thrown vertically upwards with a velocity of 49 m/s.
Calculate:
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.
🧠 In this video, you’ll learn:
✅ How to apply the three equations of motion under uniform acceleration (gravity).
✅ How to calculate maximum height (when final velocity becomes zero).
✅ How to calculate total time using initial velocity and acceleration due to gravity.
✅ How to apply the correct sign convention for upward and downward motion.
📐 Formulae Used:
1️⃣ \( v^2 = u^2 - 2gh \)
2️⃣ \( t = \frac{2u}{g} \)
Given:
u = 49 m/s, v = 0 (at maximum height), g = 9.8 m/s²
✔ Step-by-Step Calculation:
(i) Maximum height,
\( h = \frac{u^2}{2g} = \frac{49^2}{2 \times 9.8} = 122.5 \, \text{m} \)
(ii) Total time to return,
\( t = \frac{2u}{g} = \frac{2 \times 49}{9.8} = 10 \, \text{s} \)
✅ Final Answer:
• Maximum height = 122.5 m
• Total time = 10 s
📘 Based on:
Chapter 9 – Gravitation (Class 9 Science, NCERT Book)
Aligned with *CBSE 2026 Board Exam Syllabus*
📥 For full NCERT Solutions and Video Lectures of Class 9 & 10 (Maths & Science):
Download our Android App 👇
📱 https://play.google.com/store/apps/de...
📌 Like 👍 | Share 🔁 | Subscribe 🔔 for more NCERT Solution Videos
📞 Call: 999-266-1556
🌐 Visit: www.saraswatividyamandir.com
#Class9Science #Gravitation #NCERTSolutions #ExerciseQ13 #CBSE2026 #PhysicsClass9 #SaraswatiVidyamandir #ScienceWithShivangSir #EquationsOfMotion