Chapter 9 Gravitation

NCERT Class 9th

Science SOLUTIONS

Subject
Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Updated on November 1, 2025 | By Learnzy Admin | 👁️ Views: 16 students

Solution
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Initial velocity (u) = 40 m/s
Acceleration due to gravity (g) = –10 m/s² (negative because gravity acts downward)
At maximum height, final velocity (v) = 0

Using the equation of motion:
v² = u² + 2as

0 = (40)² + 2(–10)s
0 = 1600 – 20s
20s = 1600
s = 80 m

Maximum height = 80 m

When the stone returns to the ground:

Net displacement = 0 (since it comes back to the starting point)
Total distance covered = Distance up + Distance down = 80 + 80 = 160 m

Final Answers:
Maximum height = 80 m
Net displacement = 0
Total distance covered = 160 m

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