Question

A ball is throw up at speed of 4 m/s. Find the maximum height reached by the ball. Take g =10 m/s ².​ give right answer​

Answer 1

Answer:

Correct Question:

A ball is thrown up at a speed of 4m/s. Find the maximum height reached by the ball. Take g = 10m/s².

Given:-

u = 4m/s

g = 10m/s²

v = 0 m/s

where 'u' means initial velocity;

'g' means acceleration due to gravity

and 'v' means final velocity.

To find:

Maximum height reached by the ball.

Solution:

We know that,

$v {}^{2} = u {}^{2} + 2gh$

$\Longrightarrow$ 0² = 4² + 2 × 10 × h

$\Longrightarrow$ 16 = 20 h

$\Longrightarrow$ h = 16/20

= 4/5 = 0.8 m

$\Longrightarrow$ h = 0.8 m

Therefore, maximum height reached by the ball is 0.8m

Answer 2

Required Solution:

Given:

• Initial velocity (u) = 4m/s
• Final velocity (v) = 0 [As it will stop after travelling.]
• Acceleration due to gravity (g) = -10m/s²

To calculate:

• Maximum height attained by the ball (s)

Calculation:

By using the third equation of motion:

• 2as = v² - u²

→ 2 × (-10) × s = (0)² - (4)²

→ -20s = 0 - 16

→ -20s = -16

→ s = $\dfrac{-16}{-20}$

→ s = $\dfrac{4}{5}$

→ s = $0.8 \: m$⠀⠀⠀[Ans.. ]

Therefore,the maximum height reached by the ball is 0.8m.

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More information!

Equations of motion:

• v = u + at
• s = ut + ½at²
• v² – u² = 2as

Where,

• ★ v = final velocity = m/s
• ★ u = initial velocity = m/s
• ★ a = acceleration = m/s²
• ★ s = distance/displacement = m
• ★ t = time = sec

Remember that!

• When a body starts from rest, its initial velocity is 0.
• When a body comes to stop or applies breaks, its final velocity is 0.
• When the body is thrown vertically upward, then g= –10m/s².
• When the body is thrown vertically downwards, then g= 10m/s².

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