Question

3. A stone is thrown in a vertically upward direction with a vilocity of 5 m per sec if the acceleration of stone during its motion is 10m per sec square in the downward what will be the hight attend by the stone and how much time will it take to reach their​

Answer 1

Given:-

• Initial velocity ,u = 5m/s
• Final velocity ,v = 0m/s
• Acceleration due to gravity ,g= 10m/s²

To Find:-

• Maximum height attained by stone ,h
• Total time taken to reach maximum height ,t

Solution:-

⠀⠀⠀⠀⠀According to the Question

It is given that the stone is thrown vertically upward direction . So the acceleration due to gravity on stone is negative.

→ g = -10m/s²

Now, firstly we calculate 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 = 5² + 2(-10) × h

$:\implies$ 0 = 25 -20h

$:\implies$ -25 = -20h

$:\implies$ 25 = 20h

$:\implies$ 25/20 = h

$:\implies$ h = 25/2

$:\implies$ h = 1.25 m

• Hence, the maximum height attained by stone is 1.25 metres.

Now , calculating the time taken to reach maximum height .

Using 1st equation of motion

• v = u + gt

Substitute the value we get

$:\implies$ 0 = 5 + (-10) × t

$:\implies$ -5 = -10×t

$:\implies$ 5 = 10 × t

$:\implies$ 5/10 = t

$:\implies$ t = 5/10

$:\implies$ t = 0.5s

• Hence, the time taken to reach maximum height is 0.5 second .

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Answer 2

Given :-

A  stone is thrown in a vertically upward direction with a vilocity of 5 m per sec if the acceleration of stone during its motion is 10m per sec square in the downward

To Find :-

Height

Time

Solution :-

We know that

v² - u² = 2gh

0² - 5² = 2(-10)(h)

0 - 25 = -20h

-25 = -20h

-25/-20 = h

5/4 = h

1.25 = h

We know that

v = u + gt

0 = 5 + (-10)(t)

0 - 5 = -10t

-5 = -10t

-5/-10 = t

0.5 = t