Question

A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.​

Answer 1

Answer:

Time to reach Maximum height ,

t = 6/2 = 3 s.

v = 0 (at the maximum height )

a = – 9.8 m s-²

a) 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

b)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.

c) 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