Question

A ball thrown vertically upwards from the ground level hits the ground after
4seconds. Calculate the maximum height it reached during its journey.

Answer 1

Given :

▪ A ball is thrown vertically upward from the ground.

▪ Total time of flight (T) = 4s

To Find :

▪ Maximum height attained by ball.

Concept :

☞ For a body thrown vertically up with initial velocity u,

↗ Total time of flight :

$\bigstar\underline{\boxed{\bf{\purple{T=\dfrac{2u}{g}}}}}$

↗ Maximum height reached :

$\bigstar\underline{\boxed{\bf{\orange{H=\dfrac{u^2}{2g}}}}}$

Calculation :

๏ Initial velocity :

$\dashrightarrow\sf\:T=\dfrac{2u}{g}\\ \\ \dashrightarrow\sf\:4=\dfrac{2u}{10}\\ \\ \dashrightarrow\sf\:u=4\times 5\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{u=20\:mps}}}}$

๏ Maximum height :

$\Rightarrow\sf\:H=\dfrac{u^2}{2g}\\ \\ \Rightarrow\sf\:H=\dfrac{(20)^2}{2\times 10}\\ \\ \Rightarrow\underline{\boxed{\bf{\gray{H=20m}}}}$

Answer 2

Answer:

• The Maximum height (H) is 20 meters

Given:

1. Total Time period (T) = 4 Seconds.
2. Acceleration due to gravity (g) = - 10 m/s²

Explanation:

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As the question states that the Total time period (T) of the ball is 4 seconds. Then, we need to find Time of ascent (t).

From the relation we know,

⇒ T = t_{a} + t_{d}

Here,

• T Denotes Time period.
• t_{a} Denotes Time of ascent.
• t_{d} Denotes Time of descent.

Solving & Substituting the values,

⇒ T = t + t           ∵  [t_{a} = t_{b}]

⇒ T = 2 t

⇒ t = T/2

⇒ t = 4/2

⇒ t = 2

⇒ t = 2 sec

∴ We got the time of ascent.

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Now, Finding the Initial velocity of the ball,

From the formula we know,

⇒ v = u + a t

Here,

• v Denotes final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• t Denotes time period.

We know, at highest position the final velocity (v) of the ball will be zero.

Substituting the values,

⇒ 0 = u + (-10) × 2

⇒ - u = - 10 × 2

⇒ - u = - 20

⇒ u = 20

⇒ u = 20 m/s

∴ We got the Initial velocity.

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From third kinematic equation we know,

⇒ v² - u² = 2 a H

Here,

• v Denotes final velocity.
• u Denotes Initial velocity.
• a Denotes acceleration.
• H Denotes height.

Substituting the values,

⇒ (0)² - (20)² = 2 × - 10 × H

⇒ 0 - 400 = - 20 H

⇒ - 20 H = - 400

⇒ 20 H = 400

⇒ H = 400/20

⇒ H = 20

⇒ H = 20 m

∴ The Maximum height (H) is 20 meters.

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