Question

a stone tool 16second to return to its original position. Find out the initial velocity with which it was thrown up and calculate the maximum height attained by it​

Answer 1

Answer

According to the Question

Since the ball return to its initial position . So the time taken by ball to reach maximum height .

$:\implies$ t = 16/2

$:\implies$ t = 8s

Now, final velocity of stone = 0m/s (as it reaches its maximum height).

Acceleration due to gravity ,g = 10m/s²

Using first equation of motion

• v = u + gt

where,

• v denote final velocity
• u denote initial velocity
• g denote acceleration due to gravity
• t denote time taken.

Substitute the value we get

$:\implies$ 0 = u + (-10) × 8

$:\implies$ -u = -10 × 8

$:\implies$ -u = -80

$:\implies$ u = 80m/s

• Hence, the initial velocity of the stone is 80m/s.

Now, Calculating the maximum height attained by the stone

Using 3rd equation of motion

• v² = u² + 2gh

where,

• v denote final velocity
• u denote initial velocity
• g denote acceleration due to gravity
• h denote maximum height attained by stone

Substitute the value we get

$:\implies$ 0² = 80² + 2(-10) × h

$:\implies$ 0 = 6400 -20×h

$:\implies$ -6400 = -20h

$:\implies$ 6400 = 20h

$:\implies$ 20h = 6400

$:\implies$ h = 6400/20

$:\implies$ h = 320 m

• Hence, the maximum height attained by the stone is 320 metres.

Answer 2

Given :-

a stone tool 16second to return to its original position.

To Find :-

Initial velocity

Maximum height reached

Solution :-

We know that

time = 16/2 = 8 sec

g = v - u/t

-10 = 0 - u/8

-10 × 8 = 0 - u

-80 = -u

80 = u

Now

v² - u² = 2as

(0)² - (80)² = 2(-10)(s)

0 - 6400 = -20s

-6400 = -20s

-6400/-20 = s

6400/20 = s

640/2 = s

320 = s