Question

A ball is thrown vertically upwards. When it reaches half of its maximum height,
it has a velocity of 7 . What is the maximum height attained by the ball?
(g = 9.8 ms–2).

Answer 1

Answer -

Maximum Height attained by object is 5 m

Solution -

We can use the third equation of motion for the object when it reaches half of its max. height , then by multiplying by 2 we can get the maximum height attained by ball

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When ball reaches half of max. height -

$\implies$$\rm u = 7 m/s ( given )$

$\implies$$\rm v = 0 m/s$ ( because the object reaches max. height )

$\implies$$\rm a = g = - 9.8 m/s^2$ ( because object is thrown vertically upwards )

By third equation of motion -

$\implies$$\rm v^2 = u^2 \pm 2as$

$\implies$$\rm 0^2 = 7^2 - 2 \times 9.8 \times s$

$\implies$$\rm 49 = 2 \times 9.8 \times s$

$\implies$$\rm 49 = 19.6 s$

$\implies$$\rm s = \frac{49}{19.6}$

$\implies$$\rm s = 2.5 meter$

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After covering half of the max. height the ball covered the distance of 2.5 metre.

To find max. height reached we need to multiply it by 2

$\implies$$\rm Max. height = 2.5 \times 2$

$\implies$$\rm Max. height = 5 m$

Max. height reached by object is 5 m.

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Answer 2

Answer :

➥ The maximum height attained by the ball = 5 m

Given :

➤ Initial velocity of the body (u) = 7 m/s

➤ Acceleration due to gravity (g) = 9.8 m/s²

To Find :

➤ Maximum height attained by the ball (h) = ?

Required Solution :

❒ According to the given Question, given that Initial velocity of the body is 7 m/s, Acceleration due to gravity is -9.8 m/s² because ball is going against the gravity, Final velocity of of ball is 0 m/s because the object reaches max. height.

Now we have also a Acceleration due to gravity and Final velocity of a ball,

• Acceleration due to gravity (g) = -9.8 m/s²
• Final velocity of a ball (v) = 0 m/s

Note :

→ If the maximum height is h, then it's half of the maximum height is h/2.

So ,

• Maximum height of a ball = h/2

We can find maximum height attained by the ball by using the third equation of motion which says v² = u² + 2gh.

Here,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• g is the Acceleration due to gravity in m/s².
• h is the Height in m.

✎ So let's find Height (h) !

⇛ v² = u² + 2gh

⇛ 0² = 7² + 2 × (-9.8) × h/2

⇛ 0 = 49 + 2 × (-9.8) × h/2

⇛ 0 - 49 = 2 × (-9.8) × h/2

⇛ -49 = -19.6 × h/2

⇛ -49 = -9.8 × h

⇛ -49 = -9.8h

⇛ -49/-9.8 = h

⇛ 49/9.8 = h

⇛ 5 = h

⇛ h = 5 m

║Hence, the maximum height attained by the ball is 5 m.║

$\:$

Some related equations :

⪼ First equation of motion: v = u + gt

⪼ Second equation of motion: h = ut + ½ gt²

⪼ Third equation of motion: v² = u² + 2gh

Where,

• v is the Final velocity in m/s.
• u is the Initial velocity in m/s.
• f is the Acceleration due to gravity in m/s².
• t is the time taken second.
• h is the Height in m.

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