A BALL IS THROWN VERTICALLY UPWARDS WITH AN INITIAL VELOCITY OF 49 M/S. CALCULATE THE MAXIMUM HEIGHT ATTINED AND THE TIME TAKEN BY IT BEFORE IT REACHES THE GROUND AGAIN .(G=9.8M/S^2)

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Solution

Initial velocity, u = 49 m/s
At the maximum height, the velocity becomes zero.
Using, v² = u² + 2aS
=> 0 = u² – 2gh
=> h = 49²/(2×9.8)
=> h = 122.5 m
The velocity with which the ball reaches the ground is, v = -u
Using, v = u + at
=> -u = u –gt
=> t = 2u/g = 2×49/9.8 = 10 s
The ball comes back to ground in 10 s

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