A ball is released from the top of a tower of height h meter . It takes t second to reach the ground . What is the position of ball at t/3 second?
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We have, initial velocity, u = 0
Acceleration, g = 9.8 m/s2
Now,
S = ut + ½ at2
=> h = 0 + ½ (9.8)T2
=> h = ½ (9.8)T2
Again,
S = ut + ½ at2
=> S = 0 + ½ (9.8)(T/3)2
=> S = (1/9)[½(9.8)T2]
=> S = h/9
Thus, after T/3 seconds the ball is at a distance h/9 meters from the top. Or, it is (h – h/9 =) 8h/9 meters from the ground.
Acceleration, g = 9.8 m/s2
Now,
S = ut + ½ at2
=> h = 0 + ½ (9.8)T2
=> h = ½ (9.8)T2
Again,
S = ut + ½ at2
=> S = 0 + ½ (9.8)(T/3)2
=> S = (1/9)[½(9.8)T2]
=> S = h/9
Thus, after T/3 seconds the ball is at a distance h/9 meters from the top. Or, it is (h – h/9 =) 8h/9 meters from the ground.
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