5.3 Solved examples

  Motion in One Dimension

Question: A ball is projected vertically upward with a speed of $40\ m/s$. Find

(a). the maximum height
(b). the time to reach the maximum height
(c). the speed at half the maximum height

Solution:

(a). At maximum height velocity is zero.

$\overrightarrow u = + 40m/s$
$\overrightarrow a = - g = - 10m/{s^2}$
$\overrightarrow v = 0$
$\overrightarrow s = + {H_{\max }} = ?$

Kinematic equation relating velocity and displacement is,
$${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$$$0 = {\left( {40} \right)^2} + 2\left( { - 10} \right)\left( { + {H_{\max }}} \right)$$$${H_{\max }} = \frac{{1600}}{{20}}$$$${H_{\max }} = 80m$$

(b). For caluclating the time to reach the maximum height, we can write all the information below,

$\overrightarrow u = + 40m/s$
$\overrightarrow a = - g = - 10m/{s^2}$
$\overrightarrow v = 0$
$t=?$

Kinematic equation relating velocity and time as,
$$\overrightarrow v = \overrightarrow u + \overrightarrow a t$$$$0 = 40 + \left( { - 10} \right)t$$$$t = 4s$$

(c). The displacement at half the maximum height is $40\ m$ Therefore,

$\overrightarrow u = + 40m/s$
$\overrightarrow a = - g = - 10m/{s^2}$
$\overrightarrow s = +40\ m$
$\overrightarrow v = ?$

Kinematic equation relating velocity and displacement is,
$${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$$${v^2} = {\left( {40} \right)^2} + 2\left( { - 10} \right)\left( { + 40} \right)$$$${v^2} = 1600 - 800$$$${v^2} = 800$$$$v = 20\sqrt 2 m/s$$

Question: A hot-air balloon is going upwards with a velocity of $12\ m/s$. It releases a packet when it is at a height of $65\ m$ from the ground. How much time the packet will take to reach the ground. $(g=10\ m/s^2)$.

Solution:

When any particle is detached from the body then the initial velocity of the particle will be same as the velocity of the body at the time of seperation.

Therefore the velocity of the particle is $12\ m/s$.

We can write all the information in vector form as,

$\overrightarrow u = + 12\,m/s$
$\overrightarrow a = - g$
$\overrightarrow s = - 65\,m$
$t=?$

We can write the kinematic equation relating displacement and time as,
$$\overrightarrow s = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}$$$$ - 65 = 12t + \frac{1}{2}\left( { - 10} \right){t^2}$$$$ - 65 = 12t - 5{t^2}$$$$5{t^2} - 12t - 65 = 0$$$$5{t^2} - 25t + 13t - 65 = 0$$$$5t\left( {t - 5} \right) + 13\left( {t - 5} \right)$$$$\left( {t - 5} \right)\left( {5t + 13} \right) = 0$$$$t = 5s$$

Question: A falling stone takes $0.2\ s$ to fall past a window which is $1\ m$ high. From how far above the top of the window was the stone dropped?

Solution:

For position $B$.

Let the initial velocity at the top of the window is $u$

We can write all the information in vector form as ,

$\overrightarrow u = ?$
$\overrightarrow a = - g=-10\ m/s^2$
$\overrightarrow s = - 1\,m$
$t=0.2\ $

We can write the kinematic equation relating displacement and time as,
$$\overrightarrow s = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}$$$$ - 1 = u\left( {0.2} \right) + \frac{1}{2}\left( { - 10} \right){\left( {0.2} \right)^2}$$$$ - 1 = 0.2u - 0.2$$$$0.2u = - 1 + 0.2$$$$0.2u = - 0.8$$$$u = - 4\,m/s$$

For position $A$.

The initial velocity is zero as the stone was dropped from height $h$ above the top of the window.

We can write all the information in vector form as,

$\overrightarrow u = 0$
$\overrightarrow v = -4\ m/s$
$\overrightarrow a = - g=-10\ m/s^2$
$\overrightarrow s = -h=?$

We can write the equation relating velocity and displacement as,
$${v^2} = {u^2} + 2\overrightarrow a .\,\overrightarrow s $$$${\left( { - 4} \right)^2} = 0 + 2\left( { - 10} \right)\left( { - h} \right)$$$$16 = 20h$$$$h = 0.8\,m$$