A ball thrown Up vertically returns to the thrower after 6 seconds. Find i) velocity ii) Maximum height it reaches iii) Position after 4 seconds

Time to reach Maximum height ,
t = 6/2 = 3 s.

v = 0 (at the maximum height )

a = – 9.8 m s-²

i) Using, v = u + at, we get
0 = u – 9.8 × 3
or, u = 29.4 ms-¹
Thus, the velocity with which it was thrown up = 29.4ms-¹

ii) Using, 2aS = v² – u², we get
S = v²- u²/2a
= 0 – 29.4 × 29.4/- 2× 9.8
= 44.1 m
Thus, Maximum height it reaches = 44.1 m.

iii) t = 4s. In 3 s, the ball reaches the maximum height and in 1 s it falls from the top.
Distance covered in 1 s from maximum height,
S = ut + 1/2at ²
= 0 + 1/2 × 9.8 × 1
= 4.9 m

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