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 )

Answer 1

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