Physics > Motion in Two Dimension > 4.0 Projectile thrown parallel to the horizontal
Motion in Two Dimension
1.0 Introduction
2.0 Projectile motion
3.0 Ground to ground projectile motion
3.1 Maximum height
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
4.0 Projectile thrown parallel to the horizontal
5.0 Projectile on an inclined plane
6.0 Relative motion between two projectiles
4.3 Solved examples
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
Question: A particle is projected from a tower with an initial velocity of $40 \ m/s$ horizontally from a point $O$ which is $100\ m$ above the ground level. Find the time in which the particle hits the ground and its distance from the foot of the tower. ($g=10 \ m/s^2$)
Solution:
| Horizontal motion | Vertical motion |
| ${u_x} = u $ | ${u_y} = 0$ |
| ${a_x} = 0$ | ${a_y} = - g=-10$ |
Time of flight
${\overrightarrow s _y} = - H=-100$
${\overrightarrow u _y} = 0$
${\overrightarrow a _y} = - g=-10$
${\overrightarrow u _y} = 0$
${\overrightarrow a _y} = - g=-10$
$t=?$
We can write,
$${\overrightarrow s _y} = {\overrightarrow u _y}t + \frac{1}{2}{\overrightarrow a _y}{t^2}$$$$ - H = 0 + \frac{1}{2}\left( { - g} \right){t^2}$$$$ - 100 = - \frac{1}{2} \times 10 \times {t^2}$$$$t = \sqrt {20} $$$$t = 2\sqrt 5 s$$
$t = 2\sqrt 5 s$ is also known as the time of flight.
$t = 2\sqrt 5 s$ is also known as the time of flight.
Range
Horizontal distance travelled during the time of flight is known as the range.
${\overrightarrow s _x} = +R$
${\overrightarrow u _x} = 40$
${\overrightarrow a _x} = 0$
${\overrightarrow u _x} = 40$
${\overrightarrow a _x} = 0$
$t=2\sqrt 5$
We can write, $${\overrightarrow s _x} = {\overrightarrow u _x}t + \frac{1}{2}{\overrightarrow a _x}{t^2}$$$$ + R = 40 \times 2\sqrt 5 + 0$$$$R = 80\sqrt 5 \,m$$
Question: A particle is projected from the top of a tower with an initial velocity of $u$ at an angle $\theta$ with the horizontal The height of the tower is $H$. Find the time in which the particle hits the ground and its distance from the foot of the tower. Also find the maximum height reached by the particle from the ground level. ($g=10 \ m/s^2$)
Solution:
We can write the kinematic variable as,
| Horizontal motion | Vertical motion |
| ${u_x} = u \cos \theta $ | ${u_y} = u \sin \theta$ |
| ${a_x} = 0$ | ${a_y} = - g$ |
Time of flight
Time taken by the particle to reach the ground is,
${\overrightarrow s _y} = - H$
${\overrightarrow u _y} = u \sin \theta$
${\overrightarrow a _y} = - g$
${\overrightarrow u _y} = u \sin \theta$
${\overrightarrow a _y} = - g$
$t=?$
We can write,
$${{\vec s}_y} = {{\vec u}_y}t + \frac{1}{2}{{\vec a}_y}{t^2}$$$$ - H = \left( { + u\sin \theta } \right)t + \frac{1}{2}\left( { - g} \right){t^2}$$$$ - H = u\sin \theta t - \frac{{g{t^2}}}{2}$$$$g{t^2} - 2u\sin \theta t - 2H = 0$$Solving the quadratic equation we get, $$t = \frac{{2u\sin \theta \pm \sqrt {4{u^2}{{\sin }^2}\theta + 8gH} }}{{2g}}$$ Neglecting the $-ve$ time, $$t = \frac{{u\sin \theta + \sqrt {{u^2}{{\sin }^2}\theta + 2gH} }}{g}$$
Maximum height
${\overrightarrow s _y} = + h_{\max}=?$
${\overrightarrow u _y} = u \sin \theta$
${\overrightarrow a _y} = - g$
${\overrightarrow u _y} = u \sin \theta$
${\overrightarrow a _y} = - g$
${\overrightarrow v _y} = 0$
We can write, $$v_y^2 = u_y^2 + 2{\overrightarrow a _y}{\overrightarrow s _y}$$$$0 = {u^2}{\sin ^2}\theta + 2\left( { - g} \right)\left( { + {h_{\max }}} \right)$$$${h_{\max }} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$$
$h_{\max}$ is the maximum height from the top of the tower.
Maximum from the ground level is, $${H_{\max }} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} + H$$
Range
Horizontal distance travelled during the time of flight is known as range.
${\overrightarrow s _x} = +R=?$
${\overrightarrow u _x} = u \cos \theta$
${\overrightarrow a _x} = 0$
${\overrightarrow u _x} = u \cos \theta$
${\overrightarrow a _x} = 0$
$t = \frac{{u\sin \theta + \sqrt {{u^2}{{\sin }^2}\theta + 2gH} }}{g}$
We can write,
$${{\vec s}_x} = {{\vec u}_x}t + \frac{1}{2}{{\vec a}_x}{t^2}$$$$ + R = \left( { + u\cos \theta } \right)\left( {\frac{{u\sin \theta + \sqrt {{u^2}{{\sin }^2}\theta + 2gH} }}{g}} \right) + 0$$$$R = \left( {u\cos \theta } \right)\left( {\frac{{u\sin \theta + \sqrt {{u^2}{{\sin }^2}\theta + 2gH} }}{g}} \right)$$
Question: A particle is thrown downwards with velocity $u$ at an angle $\theta$ with the horizontal. Find the time of flight and range of the projectile.
Solution:
| Horizontal motion | Vertical motion |
| ${u_x} = u \cos \theta $ | ${u_y} = -u \sin \theta$ |
| ${a_x} = 0$ | ${a_y} = - g$ |
Time of flight
${\overrightarrow s _y} = - H$
${\overrightarrow u _y} = -u \sin \theta$
${\overrightarrow a _y} = - g$
${\overrightarrow u _y} = -u \sin \theta$
${\overrightarrow a _y} = - g$
$t=?$
We can write,
$${{\vec s}_y} = {{\vec u}_y}t + \frac{1}{2}{{\vec a}_y}{t^2}$$$$ - H = \left( { - u\sin \theta } \right)t + \frac{1}{2}\left( { - g} \right){t^2}$$$$ - H =- u\sin \theta t - \frac{{g{t^2}}}{2}$$$$g{t^2} + 2u\sin \theta t - 2H = 0$$Solving the quadratic equation we get, $$t = \frac{{-2u\sin \theta \pm \sqrt {4{u^2}{{\sin }^2}\theta + 8gH} }}{{2g}}$$ Neglecting the $-ve$ time, $$t = \frac{{\sqrt {{u^2}{{\sin }^2}\theta + 2gH} - u\sin \theta }}{g}$$
Range
Horizontal distance travelled during the time of flight is known as range.
${\overrightarrow s _x} = +R=?$
${\overrightarrow u _x} = u \cos \theta$
${\overrightarrow a _x} = 0$
${\overrightarrow u _x} = u \cos \theta$
${\overrightarrow a _x} = 0$
$t = \frac{{\sqrt {{u^2}{{\sin }^2}\theta + 2gH} - u\sin \theta }}{g}$
We can write,
$${{\vec s}_x} = {{\vec u}_x}t + \frac{1}{2}{{\vec a}_x}{t^2}$$$$ + R = \left( { + u\cos \theta } \right)\left( {\frac{{\sqrt {{u^2}{{\sin }^2}\theta + 2gH} - u\sin \theta }}{g}} \right) + 0$$$$R = \left( {u\cos \theta } \right)\left( {\frac{{\sqrt {{u^2}{{\sin }^2}\theta + 2gH} - u\sin \theta }}{g}} \right)$$
