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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

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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?

Numerical
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Solution

Initial velocity (u) = 40m/s

g = −10m/s2

Final velocity at maximum height (v) = 0 

Third equation of motion

v2 = u2 + 2gh

0 = 402 + 2 × (−10) × h

1600 = 20h

80m = h

Max Distance = 80m

Total distance travelled by the stone = 80 + 80

= 160m

Because the stone was thrown upwards, but it came back.

Hence, total displacement = 0

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Chapter 9: Gravitation - Exercises [Page 112]

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NCERT Science [English] Class 9
Chapter 9 Gravitation
Exercises | Q 15. | Page 112

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