Question

A bowling ball is dropped from the top of a building. If it hits the ground with a speed of 37.0 m/s, how tall was the building?​

Answer 1

${\underline {\underline {\huge {\sf { Provided \: Question: } }}}}$

A bowling ball is dropped from the top of a building. If it hits the ground with a speed of 37.0 m/s, how tall was the building?

${\underline {\underline {\huge {\sf { Required \: Answer: } }}}}$

Given:

• Final velocity (v) = 37.0m/s

To calculate:

• Height of the building

Calculation:

By the third equation of motion for freely falling bodies:

• ${\underline {\boxed {\Large {\sf {\pink { 2gh = {v}^{2} - {u}^{2} } }}}}}$

Here,

• acceleration due to gravity (g) = 10m/s²

• Final velocity (v) = 37m/s [ Given ]

• Initial velocity (u) = 0m/s [ Whenever a body is dropped from a height, its initial velocity is taken as 0 ]

• height (h) = ? ⠀⠀⠀[To be calculated]

Substitute the values:

⠀⠀⠀${\underline {\boxed {\Large {\sf {\pink { 2gh = {v}^{2} - {u}^{2} } }}}}}$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 2 \times 10 \times h = {37}^{2} - {0}^{2} }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 20 \times h = 1369 - 0 }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ 20h = 1369 }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\bf { ⇢ h = \dfrac{1369}{20} }$

⠀⠀⠀

⠀⠀⠀

⠀⠀⠀$\boxed { \bf { ⇢ h = 68.45 \: m}}$

Therefore,the height of the building is 68.45m.

$\rm \blue { So, we \: are \: done!!}$

____________________________________

Answer 2

v = + 37 m/s

u = 0 m/s

g = 10 m/s²

We know,

2gh = v² - u²

∴ h = u²/2g

⇒ h = (37)²/(20) m

⇒ h = 68.45 m.

So, the height of the structure is of 68.45 m.

More:-

Equations of motion:-

1. v = u + at
2. 2as = v² - u²
3. s = ut + ½ at²
4. Snth = u + a/2(2n - 1).