A ball is released from the top of tower of height h meter .it takes T second to reach the ground.what is the position of the ball in T/3seconds
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Let the acceleration be 'g'
Then initial velocity = 0 m/s [in T seconds]
Let the distance be 'h' m.
We know,
Equation of motion:
s = ut + 1/2 gT²
Then,
h = 1/2gT² ......(1)
h1 = 1/2gT²/9'......(2)
Substitute (1) and (2):
We'll get h1 = h9
∴ The distance from ground is h-h/9 = 8h/9 ←(Answer)
Good Studies!
Then initial velocity = 0 m/s [in T seconds]
Let the distance be 'h' m.
We know,
Equation of motion:
s = ut + 1/2 gT²
Then,
h = 1/2gT² ......(1)
h1 = 1/2gT²/9'......(2)
Substitute (1) and (2):
We'll get h1 = h9
∴ The distance from ground is h-h/9 = 8h/9 ←(Answer)
Good Studies!
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