In this page, you can find the complete solutions of the second exercise of Kinematics: The Geometry of Motion chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, Kinematics is the 22nd chapter and has three exercises only. Out of which, this is the solution of the second exercise.


Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.


Here, we have used the following notations:

u = intial velocity

a = acceleration

v = final velocity

t = time




1 i) A particle is dropped from rest. Find its velocity and position after 5 s (g = 10m/s^2).

Solution:
Here,
$u = 0 m/s$
$t = 5 s$

$g= 10 m/s^2$
To find: final velocity (v) = ?

We know,
$v = u + gt$
$or, v = 0 + 10 * 5$
$\therefore v = 50 m/s$

1 ii) A ball is projected vertically upwards with a velocity of 40 m/s. Find the velocity and position at the end of 3 s. (g = 10m/s^2)

Solution:
Here,
$u = 40 m/s$
$g = - 10 m/s^2$ [ In opposite direction of initial velocity ]

$t = 3 s$
We know,
$v = u + gt$
$or, v = 40 + (-10)3$

$or, v = 40 - 30$

$\therefore v = 10 m/s$
And,
$v^2 = u^2 + 2gh$
$or, 10^2 = 40^2 + 2(-10)h$

$or, 20 h = 1600 - 100$
$or, 20 h = 1500$
$\therefore h = 75 m$

2 i) A stone is thrown vertically upwards with an initial velocity of 20 m/s Find the height the stone can rise. (g = 10 m/s^2)

Solution:
Here,
$u = 20 m/s$
For the maximum height the stone can rise, final velocity $(v) = 0 m/s$

$g = -10m/s^2$
We know,
$v^2 = u^2 + 2gh$
$or, 0^2 = 20^2 + 2(-10)h$
$or, 20 h = 400$
$\therefore h = 20 m$

2 ii) Find the maximum height attained by a ball projected vertically upwards at the rate of 78.4 m/s. (g = 9.8 m/s^2)

Solution:
Here,
$u = 78.4 m/s$
$v = 0 m/s$ [Maximum height is attained]

$g = -9.8 m/s^2$ [ Opposite in direction with respect to u ]

Now,
$v^2 = u^2 + 2gh$
$or, 0^2 = 78.4^2 + 2(-9.8)h$
$or, 19.6h = 6146.56$
$\therefore h = 313.6 m$

3 i) A body is projected vertically upwards at 39.2 m/s. When will its velocity be 29.4 m/s? (g = 9.8 m/s^2)

Solution:

$u = 39.2 m/s$

$v = 29.4 m/s$

$g = 9.8 m/s$

To find: time taken (t) = ?

We know,

$v = u - gt$

$or, gt = u - v$

$or, t = \dfrac{u - v}{g}$

$= \dfrac{39.2 - 29.4}{9.8}$

$= 1 s$

Hence, in 1 seconds, the velocity of the body will be 29.4 seconds.

3 ii) A balloon is going vertically upwards at a speed of 20 m/s. When it was 50 m above the ground, an object is dropped form it. How long does the boject take tot reach the ground? (g = 10 m/s^2)

Solution:

Given,

$u = 20m/s$

$g = 10m/s^2$

$h = 60m$

To find: t = ?

We know,

$h = -ut + \dfrac{1}{2} * gt^2$

$or, 60 = -20t + \dfrac{1}{2} * 10t^2$

$or, 60 = -20t + 5t^2$

$or, 12 = -4t + t^2$

$or, t^2 -4t -12 = 0$

$or, (t-6)(t+2)= 0$

Either $t = 6$ or $t= -2$

Since $t \neq -2$, $t= 6s$

Hence, the object takes 6 seconds to reach the ground.

In this question, the initial velocity of the body is acting in upwards direction. But the height has to be travelled vertically downards. Also, acceleration due to gravity is acting vertically downards. So, we used a negative sign in front of initial velocity.


About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)