A ball released from the top of a tower of height h. It takes T to reach the ground. The position of the ball after T/3 is
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Answer:
The distance from ground is 8h/9
Explanation:
Let the acceleration be 'g'
Then initial velocity = 0 m/s
Let the distance be 'h' m.
time = T
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
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