3.6 Solved Examples

  Motion in Two Dimension

Question: A particle is projected with a velocity of $20\ m/s$ at an angle of $30^\circ $ with the horizontal. Find the following,

(a). Maximum height
(b). Time of flight
(c). Range

(Take $g=10\ m/s^2$)

Solution:

Maximum height

Velocity at maximum height becomes zero.

${\overrightarrow u _y} = u\sin \theta = 10$
${\overrightarrow a _y} = - g = - 10$
${\overrightarrow v _y} = 0$
${\overrightarrow s _y} = ?$

So, $$v_y^2 = u_y^2 + 2{\overrightarrow a _y}{\overrightarrow s _y}$$$$0 = {10^2} + 2\left( { - 10} \right)({H_{\max }})$$$${H_{\max }} = \frac{{100}}{{20}}$$$${H_{\max }} = 5m$$

Time of flight

When the particle returns back to the ground, the displacement in vertical direction is zero.

${\overrightarrow u _y} = u\sin \theta = 10$
${\overrightarrow a _y} = - g = - 10$
${\overrightarrow s _y} = 0$
$t=T=?$

So, kinematic equation relating displacement and time is,
$${\overrightarrow s _y} = {\overrightarrow u _y}t + \frac{1}{2}{\overrightarrow a _y}{t^2}$$$$0 = \left( {10} \right)T + \frac{1}{2}\left( { - 10} \right){T^2}$$$$T = \frac{{2 \times 10}}{{10}}$$$$T = 2s$$

Range

Horizontal distance travelled during the time of flight is range.

$${\overrightarrow s _x} = {\overrightarrow u _x}t + \frac{1}{2}{\overrightarrow a _x}{t^2}$$$$R = \left( {10\sqrt 3 } \right)2 + 0$$$$R = 20\sqrt 3 m$$

Question: A particle is projected with a velocity of $10\ m/s$ at an angle of $30^\circ $ to the horizontal. Find its distance from the point of projection $\left( {\frac{1}{2}} \right)$ seconds later. $\left( {g = 10\,m/{s^2}} \right)$

Solution:

Distance from the point of projection is,

Question: A body of mass $m$ is thrown upwards with velocity $u$ at angle $\theta$ with the horizontal. Find the velocity of the projectile after $t$ seconds.

Solution:

Horizontal motion	Vertical motion
${u_x} = u\cos \theta $	${u_y} = u\sin \theta$
${a_x} = 0$	${a_y} = - g=-10$
Velocity in $x$ direction after $t$ seconds, $${\overrightarrow v _x} = {\overrightarrow u _x} + {\overrightarrow a _x}t$$$${\overrightarrow v _x} = u\cos \theta + 0$$$${\overrightarrow v _x} = u\cos \theta $$	Velocity in $y$ direction after $t$ seconds, $${\overrightarrow v _y} = {\overrightarrow u _y} + {\overrightarrow a _y}t$$$${\overrightarrow v _y} = u\sin \theta + \left( { - g} \right)t$$$${\overrightarrow v _y} = u\sin \theta - gt$$

$$v = \sqrt {v_x^2 + v_y^2} $$$$v = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta - gt} \right)}^2}} $$$$v = \sqrt {{u^2}{{\cos }^2}\theta + {u^2}{{\sin }^2}\theta + {g^2}{t^2} - 2ugt\sin \theta } $$$$v = \sqrt {{u^2} + {g^2}{t^2} - 2ugt\sin \theta } $$

Question: A particle is projected from ground level with an initial velocity of $35\ m/s$ at angle of ${\tan ^{ - 1}}\left[ {\frac{3}{4}} \right]$ to the horizontal. Find the time for which the particle is more than $2\ m$ above the ground. $\left( {g = 10\,m/{s^2}} \right)$

Solution: Given, $$\theta = {\tan ^{ - 1}}\left[ {\frac{3}{4}} \right]\quad {\text{or}}\quad \tan \theta = \frac{3}{4}$$ So, $$\sin \theta = \frac{3}{5}\quad {\text{and}}\quad \cos \theta = \frac{4}{5}$$

When the particle travel between points $A$ and $C$, then the particle is more than $2\ m$ above the ground.

At point $A$ and $C$, the particle has vertical displacement of $2\ m$

Horizontal motion	Vertical motion
${u_x} = u\cos \theta $	${u_y} = u\sin \theta$
${a_x} = 0$	${a_y} = - g=-10$

${\overrightarrow u _y} = u\sin \theta = 35 \times \frac{3}{5} = 21\,m/s$
${\overrightarrow a _y} = - g = - 10\,m/{s^2}$
${\overrightarrow s _y} = + 2m$
$t=?$

So, we can write,
$${\overrightarrow s _y} = {\overrightarrow u _y}t + \frac{1}{2}{\overrightarrow a _y}{t^2}$$$$ + 2 = 21t + \frac{1}{2}\left( { - 10} \right){t^2}$$$$2 = 21t - 5{t^2}$$$$5{t^2} - 21t + 2 = 0$$

Solving for $t$ we get, $${t_A} = 0.098s\quad {\text{and}}\quad {t_C} = 4.102s$$

So, the time for which the particle is between $A$ and $C$ is,
$$\Delta t = {t_C} - {t_A}$$$$\Delta t = \left( {4.102 - 0.098} \right)s$$$$\Delta t = 4.00s$$