How much would a 70.0-kg person weigh on Mercury?
1 answer:
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Answer:
h =220 m
Explanation:
Given that
u = 7 m/s
Even mass will attach but this will not produce any effect on the maximum height of the ball.Because in energy conservation the effect of mass does not present.
So the final speed of the ball will be zero at the maximum height.
v² = u² - 2 g (25 + h)
0 = 7² - 2 x 10 (25 +h)
49 = 20 ( 25 +h)
49 = 500 +20 h
Here h comes out negative that is why we are taking the 70 m/s in place of 7 m/s.
0 = 70² - 2 x 10 (25 +h) ( take g =10 m/s²)
4900 = 20 ( 25 +h)
4900 = 500 +20 h
4900- 500 = 20 h
4400 = 20 h
440 = 2 h
h =220 m
