Question

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

Answer 1

Answer:

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) Velocity

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) Maximum height it reaches

Using, 2aS = v² – u², we get

S = v²- u²/2a

S = 0 – 29.4 × 29.4/(- 2× 9.8)

S = 44.1 m

Thus, Maximum height it reaches = 44.1 m.

(iii) Position after 4 seconds

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

Answer 2

Explanation:

ur answer is in the above attachment