Question

A ball thrown vertically up from the
Top of the tower with a velocity
soms reaches the ground is 6 sec. Find
the height of the tower​

Answer 1

Given:-

• Final velocity ,v = 0m/s

• Time taken ,t = 6 s

• Acceleration due to gravity ,g = 9.8m/s²

To Find:-

• Height of Tower, h

Solution:-

Firstly we calculate the initial velocity of ball.

Using 1st Equation of Motion

• v = u +at

where,

v is the final velocity

a is the acceleration

u is the initial velocity

t is the time taken

Substitute the value we get

→ 0 = u × 9.8×6

→ u = 58.8 m/s

Now, calculating the height of the building

Using 3rd Equation of Motion

• v² = u² +2ah

Substitute the value we get

→ 0² = 58.8² + 2×9.8×s

→ 0 = 3457 .44 + 19.6 ×h

→ h = 3457.44/19.6

→ h = 176 .4 m

Therefore,the height of the building is 176. 4 metres.

Answer 2

$\large{\underbrace{\underline{\sf{Question}}}}$

A ball thrown vertically up from the top of the tower with a velocity soms reaches the ground is 6 sec. Find the height of the tower.

$\large{\underbrace{\underline{\sf{Answer}}}}$

$\large{\underbrace{\sf{Given \: that}}}$

✠ Taken time = 6 seconds.

✠ Acceleration due to gravity = 9.8 m/s²

✠ Final velocity = 0 m/s

$\large{\underline{\sf{Why \: final \: velocity \: is \: 0m/s}}}$

✠ The final velocity is 0 because that time the body is at a rest that's why it becomes 0 and it's already known to us that when an object is at a rest then the velocity becomes 0 always!

$\large{\underbrace{\sf{To \: find}}}$

✠ Height of the tower.

$\large{\underbrace{\sf{Solution}}}$

✠ Height of the tower = 176.4 metres

$\large{\underbrace{\sf{We \: also \: write \: these \: as}}}$

✠ Gravity as g.

✠ Acceleration as a.

✠ Height as h.

✠ Final velocity as v

✠ Initial velocity as u.

✠ Time as t

✠ Distance as s

$\large{\underbrace{\sf{Using \: concepts}}}$

✠ Newton's first law of motion.

✠ Newton's 3rd law of motion.

$\large{\underbrace{\sf{Using \: formula}}}$

✠ Newton's first law of motion ➝ v = u + at

✠ Newton's 3rd law of motion ➝ v² = u² + 2as

$\large{\underbrace{\underline{\sf{Understanding \: the \: question}}}}$

✠ This question says that a ball thrown vertically up from the top of the tower with a velocity soms reaches the ground is 6 sec. Find the height of the tower. And it's final velocity became 0 m/s and the acceleration due to gravity = 9.8 m/s²

$\large{\underbrace{\underline{\sf{Procedure \: of \: the \: question}}}}$

✠ To solve this question 1stky we hav to use the formula of Newton's first law of motion we have to put the values then we get the initial velocity very easily. Now using Newton's third law of motion we have to put the values and we get the height of the tower very easily.

$\large{\underbrace{\sf{Full \: solution}}}$

Using formula of Newton's first law of motion we have to put the values then we get the initial velocity.

↦ v = u + at

↦ 0 = u × 9.8 × 6

↦ u = 58.8 m/s

Now, using Newton's third law of motion we have to put the values and we get the height of the tower.

↦ v² = u² + 2as

↦ 0² = 58.8² + 2(9.8)(s)

↦ 0² = 58.8² + 2 × 9.8 × s

↦ 0 = 34.57.44 + 19.6 × s

↦ s = 34.57.44 / 19.6

↦ s = 176.4 metres