Physics PROJECTILE MOTION

### Topics Covered

• Projectile Motion
• Equation of Path of a Projectile
• Time of Maximum Height
• Maximum Height of a Projectile
• Horizontal Range of a Projectile

### Projectile Motion

\color{red} ✍️ An object that is in flight after being thrown or projected is called a text(projectile). Such a projectile might be a football, a cricket ball, a baseball or any other object.

\color{red} ✍️ The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions.

• 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 vecv_o that makes an angle theta_o with the x-axis as shown in Fig.

After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward:
veca=-ghatj
Or, a_x=0, a_y=-g

color(green)("The components of initial velocity") vecv_o are:
color(green)(v_(o x)=v_o cos theta_o)

color(green)(v_(o y)=v_o sin theta_o)

If we take the initial position to be the origin of the reference frame as shown in Fig. We have,
x_o=0, y_o=0

Then, x=x_0+v_(0x)t+1/2a_xt^2

x=v_(o x)t=(v_o cos theta_o)t

and y=y_0+v_(0y)t+1/2a_yt^2

y=(v_o sin theta_o)t-1/2 g t^2

These equations give the x-, and y-coordinates of the position of a projectile at time t in terms of two parameters — initial speed v_o and projection angle theta_o.

• 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.

The components of velocity at time t can be obtained using v_x = v_(o x) + a_xt and v_y= v_(oy) a_yt.
so, v_x=v_(o x)=v_ocos theta_o

v_y=v_o sin theta_o -g t

• Note that at the point of maximum height, v_y=0, :. theta=tan^(-1)((v_y)/(v_x))=0
Q 3015591460

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 m s^(-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) ).

Solution:

We choose the origin of the x-,and yaxis 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_(o x) t
y (t) = y_o + v_(oy) 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_(o x) = 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) – g t
so that when the stone hits the ground :
v_(o x) = 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 ms^(-1)

### Equation of Path of a Projectile

As we discussed earlier, x=v_(o x)t=(v_o cos theta_o)t and y=(v_o sin theta_o)t-1/2 g t^2.

Now eliminating the time between the expressions for x and y, we obtain :

color(blue)(y=(tantheta_o) x -g/(2(v_o cos theta_o)^2) x^2)

Since g,theta_o and v_o are constants,

we can say that this equation is of the form y=ax+by^2, in which a and b are constants.

This equation represents to a color(red)(text(parabola)), i.e. color(green)("the path of the projectile is a parabola.")

### Time of Maximum Height

Let color(green)("time of maximum height"=t_m

At the point of maximum height v_y=0

we have equation, color(red)(v_y = v_o sin theta_o – g t)

So, v_y = v_o sin theta_o – g t_m=0

t_m=(v_o sin theta_o)/g

\color{fuchsia} \★mathbf\ul"Time of Flight"

The total time T_f during which the projectile is in flight can be obtained by putting y=0 in the equation

y = (v_o sin theta_o ) t – 1/2 g t^2.

color(blue)(T_f= (2(v_o sin theta_o))/g)

color{green}{T_f" is known as the time of flight of the projectile."}

• Note that color{red}{T_f = 2 t_m} (because of the symmetry of the parabolic path)
Q 3045691563

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.

Solution:

(a) The maximum height is given by

h_m = ( v_0 sintheta_0)^2/(2g) = (28 sin 30^0)^2/{2(9.8)} m

= (14xx14)/(2xx9.8) = 10.0 m

(b) The time taken to return to the same level is T_f = (2 v_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 = (v_0^2 sin 2 theta_0)/g = (28xx28xxsin60^0)/(9.8) = 69 m

### Maximum Height of a Projectile

The maximum height color(red)(h_m) reached by the projectile can be calculated by substituting t = t_m

in equation y = (v_o sin theta_o ) t – 1/2 g t^2.

y=h_m=(v_o sin theta_o)((v_o sin theta_o)/g)-g/2 ((v_o sin theta_o)/g)^2

color(red)(h_m=(v_o sin theta_o)^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
color(red)(R=(v_o cos theta_o) (t_f))

color(blue)(R=(v_o cos theta_o) ( 2 v_o sin theta_o) / g)

color(red)(R=(v_o^2 sin 2theta_o)/g)

• R is maximum when sin 2theta_o is maximum, i.e., when theta_o = 45^o

color(green)(ul"The maximum horizontal range is" R_m=(v_o^2)/g)
Q 3055491364

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.

Solution:

For a projectile launched with velocity v_o at an angle θ_o , the range is given by

R = (v_0^2 sin2 theta_0)/g

Now, for angles, (45° + α) and ( 45° – α), 2θ_o 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.