Question 15 A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s 2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?Class 9 - Science - Gravitation Page 70
Answers
Answered by
0
Maximum height = V^2/2g= 80m
Net displacement zero assuming the stone falls back in ground.
Total distance equals twice of maximum height.
= 160m.
Net displacement zero assuming the stone falls back in ground.
Total distance equals twice of maximum height.
= 160m.
Answered by
0
Hi friend,
Maximum height = v²/2g
Height = 160/20
Maximum height = 8m
Displacement of the stone will be zero as the body comes back to its original position.
Total distance = 2× height
Distance =2×80
Distance = 160m.
Hope this helped you a little!!!
Maximum height = v²/2g
Height = 160/20
Maximum height = 8m
Displacement of the stone will be zero as the body comes back to its original position.
Total distance = 2× height
Distance =2×80
Distance = 160m.
Hope this helped you a little!!!
Similar questions