Question

A ball thrown up vertically returns to the thrower after 4 s. Find :
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches.
(c) its position after 3 s.

Answer 1

Explanation:

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) Here, 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

Therefore, The ball will be 4.9 m below the top of the tower after 4 s.

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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) 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