As an application of the ideas developed in the previous sections, we consider the motion of a projectile. An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. It was Galileo who first stated this independency of the horizontal and the vertical components of projectile motion in his Dialogue on the great world systems (1632).
In our discussion, we shall assume that the air resistance has negligible effect on the motion of the projectile. Suppose that the projectile is launched with velocity v$_o$ that makes an angle θ$_o$ with the x-axis as shown in Fig. 1.
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| Fig.1: Motion of an object projected with velocity v$_o$ at angle θ$_0$. |
After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward:
$\mathbf{a}=-g \mathbf{\hat{j}}$
Or, $a_x$ = 0, $a_y=-g$
The components of initial velocity v$_o$ are:
$v_{0x}=v_0 \ cos \ \theta_0$
$v_{0y}=v_0 \ sin \ \theta_0$
If we take the initial position to be the origin of the reference frame as shown in Fig. 1, we have:
x$_0$ = 0, y$_0$ = 0
Then, Eq.(4.34b) becomes:
$x = v_{0x}t=(v_0 \ cos \ \theta_0)t$
and $y = (v_0 \ sin \ \theta_0)t - \frac{1}{2}gt^2$
The components of velocity at time t can be obtained using Eq.(4.33b):
$v_x=v_{0x}=v_0 \cos \ \theta_0$
Equation (4.38) gives the x-, and y-coordinates of the position of a projectile at time t in terms of two parameters — initial speed vo and projection angle θ$_o$ . Notice that the choice of mutually perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in a simplification. One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y- component changes, like an object in free fall in vertical direction. This is shown graphically at few instants in Fig. 4.18. Note that at the point of maximum height, v$_y$ = 0 and therefore,
θ = tan$^{-1}\left(\frac{v_y}{v_x}\right)$ = 0
Equation of path of a projectile
What is the shape of the path followed by the projectile? This can be seen by eliminating the time between the expressions for x and y as given in Eq. (4.38). We obtain:
$y=(tan \ \theta_0)t-\frac{g}{2(v_0 cos \ \theta_0)^2}x^2$
Now, since g, θ$_o$ and v$_o$ are constants, Eq. (4.40) is of the form y = ax + bx$^2$, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 2).
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| Fig.2: The path of a projectile is a parabola. |
Time of maximum height How much time does the projectile take to reach the maximum height? Let this time be denoted by tm. Since at this point, v$_y$= 0, we have from Eq. (4.39):
$v_y=v_0 \ sin \ \theta_0 - gt_m = 0$
Or, $t_m=\frac{v_0 sin \ \theta_0}{g}$
The total time T$_f$ during which the projectile is in flight can be obtained by putting y = 0 in Eq. (4.38). We get:
$T_f=\frac{2v_0 sin \ \theta_0}{g}$
T$_f$ is known as the time of flight of the projectile. We note that T$_f$ = 2t$_m$, which is expected because of the symmetry of the parabolic path.
Maximum height of a projectile
The maximum height hm reached by the projectile can be calculated by substituting t = t$_m$ in Eq. (4.38) :
$y=h_m=(v_0 sin \ \theta_0)\left(\frac{v_0 \ \sin \ \theta_0}{g}\right)-\frac{g}{2}\left(\frac{v_0 \ sin \ \theta_0}{g}\right)^2$
Or, $h_m=\frac{(v_0 \ sin \ \theta_0)^2}{2g}$
Horizontal range of a projectile
The horizontal distance travelled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight T$_f$. Therefore, the range R is
$R=(v_0 \ cos \ \theta_0)T_f$
$R=(v_0 \ cos \ \theta_0)(2v_0 \ sin \ \theta_0)/g$
Or, $R=\frac{v_0^2 \ sin \ 2\theta_0}{g}$
Equation (4.43a) shows that for a given projection velocity v$_o$ , R is maximum when sin 2θ$_0$ is maximum, i.e., when θ$_0$ = 45$^0$. The maximum horizontal range is, therefore,
$R_m=\frac{v_0^2}{g}$
Example 1
Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”. Prove this statement.
Answer:
For a projectile launched with velocity vo at an angle θ$_o$, the range is given by
$R=\frac{v_0^2 \ sin \ 2\theta_0}{g}$
Now, for angles, (45° + α) and (45° – α), 2θo is (90° + 2α) and (90° – 2α), respectively. The values of sin (90° + 2α) and sin (90° – 2α) are the same, equal to that of cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.
Example 2
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 ms$^{-1}$. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. (Take g = 9.8 m.s$^{-2}$).
Answer
We choose the origin of the x-,and y-axis at the edge of the cliff and t = 0 s at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction. The x-, and ycomponents of the motion can be treated independently. The equations of motion are:
x(t) = x$_o$ + v$_{ox}$t
y (t) = yo + voy t +(1/2)a$_y$ t$^2$
Here, x$_o$ = y$_o$ = 0, v$_{oy}$ = 0, a$_y$ = –g = –9.8 m.s$^{-2}$,
v$_{ox}$ = 15 m.s$^{-1}$.
The stone hits the ground when y(t) = – 490 m.
– 490 m = –(1/2)(9.8)t$^2$.
This gives t =10 s.
The velocity components are v$_x$ = v$_{ox}$ and v$_y$ = v$_{oy}$ – gt
so that when the stone hits the ground:
v$_{ox}$ = 15 m.s$^{-1}$
v$_{oy}$ = 0 – 9.8 × 10 = – 98 m.s$^{-1}$
Therefore, the speed of the stone is
$\sqrt{v_x^2+v_y^2}=\sqrt{15^2+98^2}$ = 99 m/s
Example 3
A cricket ball is thrown at a speed of 28 m s$^{–1}$ in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.
Answer
(a) The maximum height is given by
$h_m=\frac{(v_0 \ sin \ \theta_0)^2}{2g}$
$=\frac{(28 \ sin \ 30^0)^2}{2(9.8)}$
$h_m=\frac{14^2}{2 \times 9.8}$ = 10 m
(b) The time taken to return to the same level is
T$_f$ = (2v$_o$ sin θ$_o$)/g = (2× 28 × sin 30°)/9.8 = 28/9.8 s = 2.9 s
(c) The distance from the thrower to the point where the ball returns to the same level is
$R=\frac{v_0^2 \ sin \ 2\theta_0}{g}=\frac{28^2 \ sin \ 60^0}{9.8}$ = 69 m
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