Question

A person sitting in the ground floor of a building

notices through the window, of height 1.5 m, a ball

dropped from the roof of the building crosses the

window in 0.1 s. What is the velocity of the ball

when it is at the topmost point of the window ?

(g = 10 m/s

2

)​

Answer 1

Answer:

....................

Answer 2

Given:

A person sitting in the ground floor of a building

notices through the window, of height 1.5 m, a ball

dropped from the roof of the building crosses the

window in 0.1 s.

To find:

Initial velocity at the top of the window.

Calculation:

Let the initial velocity at the top of the window be denoted as "u".

Applying 2nd equation of kinematics:

$\therefore \: s = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: s = ut + \dfrac{1}{2} g {t}^{2}$

$= > \: 1.5 = u(0.1) + \dfrac{1}{2} (10) {(0.1)}^{2}$

$= > \: 1.5 = u( {10}^{ - 1}) + \dfrac{1}{2} (10) ( {10}^{ - 2} )$

$= > \: 1.5 = u( {10}^{ - 1}) + \dfrac{1}{2}( {10}^{ - 1} )$

$= > \: 1.5 = u( {10}^{ - 1}) + 0.05$

$= > \: u( {10}^{ - 1}) = 1.5 - 0.05$

$= > \: u( {10}^{ - 1}) = 1.45$

$= > \: u = 1.45 \times 10$

$= > \: u = 14.5 \: m {s}^{ - 1}$

So, Velocity at top of window is 14.5 m/s.