a stone is thrown vertically upward with an initial velocity of 19.6 metre per second what is the total distance covered by the stone
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Use 3rd equation of motion
v^2 = u^2 + 2gh
For upward journey
Final velocity, v = 0
Initial velocity , u = 19.6 m/s
0 = 19.6^2 - 2x9.8xh Use g =9.8
19.6^2 = 2x9.8xh
On solving h = 19.6 m
Thus distance covered during upward journey is 19.6m
Distance covered in entire journey upward + downward = 19.6x2 = 39.2m
v^2 = u^2 + 2gh
For upward journey
Final velocity, v = 0
Initial velocity , u = 19.6 m/s
0 = 19.6^2 - 2x9.8xh Use g =9.8
19.6^2 = 2x9.8xh
On solving h = 19.6 m
Thus distance covered during upward journey is 19.6m
Distance covered in entire journey upward + downward = 19.6x2 = 39.2m
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