a stone is thrown vertically upward with initial velocity 14 metre per second find maximum height reached by the stone what is the net displacement and the total distance covered by the stone
Farzeen1:
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u= 14 m/s
v= 0 m/s (since we need to find maximum height)
a= -9.8 m/s^2
s= (v^2 - u^2)/2a
= (0-196)/-19.6 = -196/19.6 = 10m
Thus the maximum height reached by the stone is 10m.
When the stone is thrown upwards, it will come back to the same place.
Thus,
Displacement= 0
Distance Covered= (10 x 2)m = 20m
v= 0 m/s (since we need to find maximum height)
a= -9.8 m/s^2
s= (v^2 - u^2)/2a
= (0-196)/-19.6 = -196/19.6 = 10m
Thus the maximum height reached by the stone is 10m.
When the stone is thrown upwards, it will come back to the same place.
Thus,
Displacement= 0
Distance Covered= (10 x 2)m = 20m
Thanks
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