what will be the velocity and acceleration of a body, dropped from a certain point at the end of 3 seconds ? (g= 9.8 m/s^2)
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From the first equation of kinematics,
v = u + at
v = final velocity
u = initial velocity = 0 here.
a = acceleration ( g here) = 9.8 m/s^2
t = time after t = 0 of the analysis of the system.
So,
v = u + gt = gt = (9.8)(3) = 28.4 m/s
This will be the velocity of the body dropped and reaching the ground
and acceleration of the body will still be 9.8 m/s^2 because the gravitational force is the only force contributing to the net force in the vertical direction when analysed for the body from the ground frame of reference.
v = u + at
v = final velocity
u = initial velocity = 0 here.
a = acceleration ( g here) = 9.8 m/s^2
t = time after t = 0 of the analysis of the system.
So,
v = u + gt = gt = (9.8)(3) = 28.4 m/s
This will be the velocity of the body dropped and reaching the ground
and acceleration of the body will still be 9.8 m/s^2 because the gravitational force is the only force contributing to the net force in the vertical direction when analysed for the body from the ground frame of reference.
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