Question

A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m ​ 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction and lands on the ground 5.0\, \text m5.0m5, point, 0, start text, m, end text below. We can ignore air resistance. What is the final velocity of the book in \dfrac{\text m}{\text s} s m ​ start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?

Answer 1

Correct Question:

A book is thrown downward from the library window with a speed of 2.0 m/s and lands on the ground 5 m below the window. What is the final velocity of the book ? Ignore air resistance.

Calculation:

Let the final velocity of the book be "v":

Now, since we are ignoring air resistance and considering gravitational acceleration to be constant, we can apply the equation of kinematics:

• Let initial velocity be u , gravitational acceleration be "g" and height be "h".

Applying 3rd equation of kinematics:

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$\implies \: {v}^{2} = {u}^{2} + 2gh$

$\implies \: {v}^{2} = {(2)}^{2} + (2 \times 10 \times 5)$

$\implies \: {v}^{2} = 4+ 100$

$\implies \: {v}^{2} = 104$

$\implies \: v = \sqrt{104}$

$\implies \: v = 10.19 \: m {s}^{ - 1}$

So, the book lands on the ground with a final velocity of 10.19 m/s.