Question

A ball thrown up vertically returns after 6s. Find:-
(i) The velocity with which it was throw up
(ii) the maximum height it reaches, and
(iii) It's position after 4s.​

Answer 1

Answer:

I had tried it maybe correct sorry for not being clear about this

Answer 2

Answer:

Time to reach Maximum height,

t = 6/2 = 3 s.

v = 0 (at the maximum height)

a = – 9.8 m s–²

Since the ball is thrown upwards, the acceleration is negative.

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

S = 0 + 1/2 × 9.8 × 1

S = 4.9 m

Explanation: