A body is dropeed from a certain height and the distance covered by it in time t second is given by S=1/2gt 2 .Find the velocity and acceleraton of the body at the end of 3 s.Take g=9.8m/s
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here,
t = 3 sec
as, s = 0.5*g*t^2 --------(i)
= 4.9*9
s = 44.1 metres
by second law of motion;
s = ut + 0.5*g*t^2 -------(ii)
from (i) and (ii),
u = 0 m/s
from third law of motion;
v² - u² = 2*a*s
v² = 2*9.8*44.1
v² = 864.36
v = √864.36
= 29.4 m/s
throughout the fall, the acceleration is g = 9.8 m/s²
therefore at t = 3 sec,
s = 44.1 m
v = 29.4 m/s
a = 9.8 m/s²
t = 3 sec
as, s = 0.5*g*t^2 --------(i)
= 4.9*9
s = 44.1 metres
by second law of motion;
s = ut + 0.5*g*t^2 -------(ii)
from (i) and (ii),
u = 0 m/s
from third law of motion;
v² - u² = 2*a*s
v² = 2*9.8*44.1
v² = 864.36
v = √864.36
= 29.4 m/s
throughout the fall, the acceleration is g = 9.8 m/s²
therefore at t = 3 sec,
s = 44.1 m
v = 29.4 m/s
a = 9.8 m/s²
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