Please Wait... While Loading Full Video#### CLASS 11 CHAPTER 4 - MOTION IN A PLANE

### PROJECTILE MOTION

• Projectile Motion

• Equation of Path of a Projectile

• Time of Maximum Height

• Maximum Height of a Projectile

• Horizontal Range of a Projectile

• Equation of Path of a Projectile

• Time of Maximum Height

• Maximum Height of a Projectile

• Horizontal Range of a Projectile

`\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`

`\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)` ).

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)`

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.")`

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.")`

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)

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.

(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`

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))`

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))`

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)`

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.

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`.