Question

$\huge \bf Question$

● A stone is thrown in a vertically upward direction with a velocity of 5ms‐¹. If the acceleration of the stone during its motion is 10ms‐² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there ?

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Answer 1

Answer:

$\huge\colorbox{yellow}{Given\:-}$

• Initial velocity (u) = 5 ms-¹

• Acceleration = 10 ms-²

$\huge\colorbox{yellow}{To\:Find\:-}$

• The height attend by the stone

• The time taken by stone to reach upward

$\huge\colorbox{yellow}{Solution\:-}$

$\Large{\red{ {v}^{2} - {u}^{2} = 2as}}$

$s = \frac{ {v}^{2} - {u}^{2} }{2as}$

$\frac{ {(0)}^{2} - {(5)}^{2} }{2 \times ( - 10)} = 1.25 \: m$

Now,

$\Large{\red{v = u + at}}$

$0 = 5 + ( - 10)t$

$t = \frac{5}{10} = 0.5 \: s$

Hence, we can conclude that

$\boxed{\pink{ = > i) \: \: \: 1.25 \: m}}$

$\boxed{\pink{ = > ii) \: \: \: 0.5 \: s}}$

$\huge\colorbox{yellow}{Thank\:You}$

Answer 2

⇒ Given:

A stone is thrown vertically upwards with a velocity of 5 m/s.

The acceleration of the stone during its downward journey is 10 m/s².

⇒ To Find:

The height attained by the stone.

The time taken.

⇒ Formulae to be used:

The first equation of motion → v = u + at

The third equation of motion → v² - u² = 2as

⇒ Solution:

Initial velocity [u] = 5 m/s

Acceleration [a] = -10 m/s² (The stone is coming downwards)

Final velocity [v] = 0 (As the stone reaches a certain height)

We can find the time:

Using the first equation of motion:

v = u + at

Substituting the values:

0 = 5 - 10 x t

$\sf{-5=-10t}$

$\sf{t=\dfrac{-5}{-10}}$

$\sf{t=0.5\:seconds}$

Hence the time taken is 0.5 seconds.

Now, we can find the height:

Using the third equation of motion:

v² - u² = 2as

Substituting the values:

0² - (5)² = 2 x -10 x s

-25 = -20s

$\sf{\dfrac{-25}{-20}=s}$

s = 1.25 m

Hence the stone will reach a height of 1.25 m.

⇒ Final Answers:

Time taken = 0.5 seconds

Height the stone reaches = 1.25 m