A stone released feom top of a tower of height 44.1 m .with wat velocity and in how much tym does it reach eart when g= 9.8 m/s sq.
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WE HAVE,
u = 0
g = 9.8
h = 44.1
v² - u² = 2*g*h
v² - 0 = 2 * 9.8 * 44.1
v² = 864.36 m/s
v = √864.36
v = 29.4 m/s
v = u + g*t
29.4 = 0 + 9.8 * t
29.4 / 9.8 = t
THEREFORE ,
t = 3 seconds
BRAINLIEST PLEASEEEE
u = 0
g = 9.8
h = 44.1
v² - u² = 2*g*h
v² - 0 = 2 * 9.8 * 44.1
v² = 864.36 m/s
v = √864.36
v = 29.4 m/s
v = u + g*t
29.4 = 0 + 9.8 * t
29.4 / 9.8 = t
THEREFORE ,
t = 3 seconds
BRAINLIEST PLEASEEEE
ecbelango:
BRAINLIEST PLSSSSSSSS
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