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Question
a stone is thrown vertically upward with an initial velocity of 30 ms-1.find out the maximum height reached by the stone. what will be the net displacements and the total distance covered by the stone?(g =10ms-2 )
a stone is thrown vertically upward with an initial velocity of 30 ms-1.find out the maximum height reached by the stone. what will be the net displacements and the total distance covered by the stone?(g =10ms-2 )
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Solution
Here,
u = 30 ms-1
g = 10 ms-2
At the top most position the velocity of the stone = 0
v = 0
v = u – gt
=> 0 = 30 – 10t
=> t = 3 s
h = ut – ½ gt2
=> h = 30×3 – ½×10×(3)2
=> h = 45 m
Max. height attained = 45 m
Since the stones will fall back to earth displacement = 0
Total distance covered = 45+45 = 90 m
Here,
u = 30 ms-1
g = 10 ms-2
At the top most position the velocity of the stone = 0
v = 0
v = u – gt
=> 0 = 30 – 10t
=> t = 3 s
h = ut – ½ gt2
=> h = 30×3 – ½×10×(3)2
=> h = 45 m
Max. height attained = 45 m
Since the stones will fall back to earth displacement = 0
Total distance covered = 45+45 = 90 m
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