JEE Physics Motion In A Plane Notes Set 07

What a Projectile Is

A projectile is any body that is launched at an angle and afterwards moves only under a constant force; in these notes that force is the body's weight. A cricket ball hit for a six and a stone thrown across a field are projectiles. A rocket or an aeroplane is not, because its engine keeps pushing it.
Three simplifications are made: the range is small enough for gravity to be uniform in size and direction, air drag is ignored, and the Earth's surface is treated as flat. With these the motion is two-dimensional with constant acceleration \( \vec a=-g\hat j \).

The key idea is independence of perpendicular motions. Horizontally nothing acts, so the velocity stays constant. Vertically the body behaves like a freely moving particle with acceleration \( g \) downward. The curved path is just these two simple motions run together in the same time, and the path turns out to be a parabola.

Projection at an Angle from Level Ground

Launch at speed \( u \) and angle \( \theta \) above the horizontal, with the origin at the launch point. The components are \( u_x=u\cos\theta \) and \( u_y=u\sin\theta \). After time \( t \): \( v_x=u\cos\theta \), \( v_y=u\sin\theta-gt \), \( x=u\cos\theta\,t \) and \( y=u\sin\theta\,t-\tfrac12gt^2 \). The speed is \( v=\sqrt{v_x^2+v_y^2} \) and its direction satisfies \( \tan\phi=v_y/v_x \).

If the landing point is at the launch level, three results follow.
1. Time of flight: the vertical displacement is zero again, so \( T=\frac{2u\sin\theta}{g} \).
2. Maximum height: the vertical velocity is zero at the top, so \( H=\frac{u^2\sin^2\theta}{2g} \).
3. Range: horizontal speed times flight time, \( R=\frac{u^2\sin2\theta}{g}=\frac{2u_xu_y}{g} \).
These three formulas hold only when the body lands at the height it started from.

Useful consequences: the range is greatest at \( \theta=45^\circ \), where \( R_{max}=u^2/g \) and \( H=R_{max}/4 \). Angles \( \theta \) and \( 90^\circ-\theta \) give equal ranges because \( \sin2\theta \) is the same, but different heights and times, and \( T_1T_2=2R/g \). The ratio \( R/H=4\cot\theta \), so range equals height when \( \tan\theta=4 \). At the highest point the velocity is purely horizontal and equals \( u\cos\theta \), while the acceleration is still \( g \) downward. Firing in all directions with speed \( u \), the ground reached is a circle of radius \( u^2/g \) and area \( \pi u^4/g^2 \).

Equation of the Path

Eliminating time between the two position equations gives \( y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta} \). It can also be written as \( y=x\tan\theta\left(1-\frac xR\right) \) or \( y=x\tan\theta-\frac{gx^2}{2u^2}(1+\tan^2\theta) \). The last form is a quadratic in \( \tan\theta \), so it tells directly how many launch angles reach a given point: two, one or none, depending on the discriminant. The coefficient of \( x^2 \) gives the horizontal speed, and the coefficient of \( x \) gives the launch slope.

Projection from a Height

Horizontal launch from height \( h \): the vertical motion starts from rest, so \( t=\sqrt{2h/g} \), the horizontal range is \( R=u\sqrt{2h/g} \), and the path is \( y=\frac{gx^2}{2u^2} \) measured downward. At landing \( v_x=u \) and \( v_y=\sqrt{2gh} \), so \( v=\sqrt{u^2+2gh} \).
Launch from a tower at angle \( \alpha \) above or below the horizontal: write the vertical equation with displacement \( -h \) and solve the quadratic for the flight time; then \( R=u\cos\alpha\,T \). The landing speed is always \( \sqrt{u^2+2gh} \) whatever the direction, since only the height dropped matters for kinetic energy. The flight time, however, is shortest for a downward throw and longest for an upward one.

Projection from a Moving Platform

Add the platform velocity to the launch velocity to get the ground-frame launch velocity. From a truck moving horizontally at \( v \), a ball thrown up straight rises and falls vertically for the person on the truck but traces a parabola for a roadside observer. Thrown forward at angle \( \alpha \) it starts with \( u_x=u\cos\alpha+v \), thrown backward with \( u\cos\alpha-v \); the vertical component is \( u\sin\alpha \) in both. From a platform rising at \( v \) the vertical component becomes \( u\sin\alpha+v \), and from one descending at \( v \) it is \( u\sin\alpha-v \). If the platform accelerates, the ball no longer shares its motion and falls behind by the extra distance the platform gains.

Projection on an Inclined Plane

Choose the \( x \)-axis along the plane and the \( y \)-axis perpendicular to it. For an incline of angle \( \beta \) and projection angle \( \alpha \) measured from the plane, up the slope: \( a_x=-g\sin\beta \), \( a_y=-g\cos\beta \), \( u_x=u\cos\alpha \), \( u_y=u\sin\alpha \). The body lands when \( y=0 \), so \( T=\frac{2u\sin\alpha}{g\cos\beta} \). The greatest distance from the plane is \( H=\frac{u^2\sin^2\alpha}{2g\cos\beta} \), and the range along the plane is \( R=\frac{2u^2\sin\alpha\cos(\alpha+\beta)}{g\cos^2\beta} \). For a launch down the slope replace \( \beta \) by \( -\beta \).
The range is largest when the launch direction bisects the angle between the plane and the vertical; then \( R_{max}=\frac{u^2}{g(1+\sin\beta)} \) up the slope and \( \frac{u^2}{g(1-\sin\beta)} \) down it. Special strikes: the body hits the plane at right angles when \( 2\tan\alpha=\cot\beta \) (its velocity along the plane has become zero), and it hits moving horizontally when it is at the top of its path in the ground frame. A body projected horizontally with speed \( u \) from the top of a plane lands at distance \( \frac{2u^2\sin\beta}{g\cos^2\beta} \) along the plane.

Bounce off a Smooth Wall

In an elastic collision with a smooth vertical wall the horizontal velocity reverses and the vertical one is unchanged, so the time of flight and maximum height are the same as without the wall. If the wall is at distance \( x \) and the range without the wall is \( R \), the body lands \( R-x \) in front of the wall, so its distance from the launch point is \( |R-2x| \): \( 2x-R \) beyond the launch point if \( x\ge R/2 \) and \( R-2x \) on the far side if the wall is nearer.

Worked Examples

Example 1. A body is thrown at 20 m/s at \( 30^\circ \) to the horizontal (\( g=10\ \text{m/s}^2 \)). \( T=\frac{2\times20\times0.5}{10}=2 \) s, \( H=\frac{400\times0.25}{20}=5 \) m and \( R=\frac{400\sin60^\circ}{10}=20\sqrt3\approx34.6 \) m.

Example 2. Find the time after which a projectile thrown at 100 m/s at \( 60^\circ \) first moves at \( 45^\circ \) to the horizontal. \( v_x=50 \) and \( v_y=50\sqrt3-10t \). Setting \( v_y=v_x \) gives \( t=5(\sqrt3-1)\approx3.66 \) s.

Example 3. A ball is fired with speed 20 m/s at \( 37^\circ \) from the ground towards a wall 18.4 m away, and rebounds elastically. \( R=\frac{400\times2(0.6)(0.8)}{10}=38.4 \) m, so it lands \( 38.4-18.4=20 \) m in front of the wall, which is \( |38.4-36.8|=1.6 \) m from the launch point.

Example 4. A stone is thrown at 10 m/s at \( 37^\circ \) above the horizontal from the top of an 11 m tower. \( u_x=8,\ u_y=6 \). Landing: \( -11=6t-5t^2 \) gives \( t=2.2 \) s, so \( R=17.6 \) m, and the landing speed is \( \sqrt{100+2\times10\times11}=\sqrt{320}=8\sqrt5 \) m/s.

Example 5. A ball rolls off a table 0.8 m high at 3 m/s. \( t=\sqrt{2(0.8)/10}=0.4 \) s, \( R=1.2 \) m, \( v_y=4 \) m/s and the landing speed is \( \sqrt{9+16}=5 \) m/s at \( \tan^{-1}(4/3) \) below the horizontal.