A fielder at the boundary throws the ball to the wicket-keeper. The ball rises, slows in its climb, passes a highest point, and comes down faster and faster. Seen from the side, the ball moves across the ground at a steady pace the whole time, even while it is rising and falling.
A projectile is two motions
A projectile is a body thrown into the air and then moving under gravity alone. Air resistance is neglected. Let the body be launched from the ground with speed \(u\) at an angle \(\theta\) above the horizontal. This angle is the angle of projection. Split the initial velocity into two components.
\[ u_x = u\cos\theta, \qquad u_y = u\sin\theta \]
Now apply the independence of motions from the last lesson. Horizontally there is no acceleration, so \(v_x = u\cos\theta\) stays constant for the whole flight. Vertically the acceleration is \(-g\) with upward positive, so the vertical motion is exactly the free fall of the last chapter, starting with velocity \(u\sin\theta\).
Watch the two components in Fig 4.1. The horizontal component never changes. The vertical component behaves exactly like the velocity of a ball thrown straight up. At the top of the path the vertical component is zero, but the ball is still moving, horizontally, at \(u\cos\theta\). The speed at the top is the smallest speed of the whole flight, but it is not zero.
At the highest point of a projectile's path, which statement is correct?
Time of flight
The projectile lands when it returns to \(y = 0\). Put \(y = 0\) in the vertical equation and solve for \(t\). One root is \(t = 0\), the launch. The other root is the time of flight.
\[ y = (u\sin\theta)\,t - \tfrac{1}{2} g t^2 = 0 \]
Why. the vertical position is zero at launch and again at landing.
\[ t\left(u\sin\theta - \tfrac{1}{2} g t\right) = 0 \]
Why. take out the common factor \(t\). One root, \(t = 0\), is the launch.
\[ T = \dfrac{2u\sin\theta}{g} \]
Read it. the other root is the landing. It is twice the time to the top, \(u\sin\theta / g\), by the symmetry of free fall.
Maximum height
At the top the vertical velocity is zero. This is the free-fall result of the last chapter with \(u\) replaced by the vertical component \(u\sin\theta\).
\[ H = \dfrac{(u\sin\theta)^2}{2g} = \dfrac{u^2\sin^2\theta}{2g} \]
The maximum height depends on the vertical component only. A larger angle gives a larger vertical component and a greater height.
Range
The range \(R\) is the horizontal distance covered in the time of flight. The horizontal velocity is constant, so \(R = u_x T\).
\[ R = (u\cos\theta)\,T = (u\cos\theta)\dfrac{2u\sin\theta}{g} \]
Why. the horizontal velocity is constant, so the range is that velocity times the time of flight.
?
Your turn. collect the factors. Which line is it?
\[ R = \dfrac{u^2\sin 2\theta}{g} \]
Read it. the identity \(2\sin\theta\cos\theta = \sin 2\theta\) turns the product into one sine.
A fielder throws the ball at \(20\,\text{m/s}\) at \(37°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\), \(\cos 37° = 0.8\). Find the time of flight, the maximum height and the range.
\(u_x = 20 \times 0.8 = 16\,\text{m/s}\). \(u_y = 20 \times 0.6 = 12\,\text{m/s}\). Horizontal: constant \(16\,\text{m/s}\). Vertical: free fall starting at \(12\,\text{m/s}\) upward.
\(T = 2u_y / g\), \(H = u_y^2 / 2g\), \(R = u_x T\).
\(T = 24 / 10 = 2.4\,\text{s}\). \(H = 144 / 20 = 7.2\,\text{m}\). \(R = 16 \times 2.4 = 38.4\,\text{m}\).
The ball is in the air for \(2.4\,\text{s}\), rises \(7.2\,\text{m}\), and lands \(38.4\,\text{m}\) away. It passes the top of its path after \(1.2\,\text{s}\), moving horizontally at \(16\,\text{m/s}\).
A projectile is launched at \(20\,\text{m/s}\) at \(30°\) above the horizontal. Take \(g = 10\,\text{m/s}^2\). Find the time of flight in seconds.
The launch speed of a projectile is doubled and the angle is kept the same. What happens to the range?
The water leaving a hose is a stream of projectiles. Each drop follows the same parabola, so the stream itself draws the trajectory in the air. A gardener who wants to reach a far corner raises the nozzle towards \(45°\). To reach over a wall, the gardener raises it further, gaining height and losing range, exactly as the formulas for \(H\) and \(R\) say. The fountains in a garden are the same physics, with the nozzle fixed.
- Putting \(v = 0\) at the highest point.Only \(v_y = 0\) at the top. The horizontal velocity \(u\cos\theta\) is unchanged, and the speed there is \(u\cos\theta\).
- Using \(u\) instead of \(u\sin\theta\) in the free-fall results.The vertical motion starts with the vertical component only. Replace \(u\) by \(u\sin\theta\) in \(T\) and \(H\).
- Forgetting that \(R\) grows as \(u^2\).The range is horizontal velocity times time of flight, and both are proportional to \(u\).