06Classical Mechanics
A ball thrown at an angle follows a parabola, but nothing new is needed to describe it: horizontally it moves with constant velocity and vertically it is a vertical throw, both happening at the same time. The two motions only share the time \(t\). Give the launch speed, the angle, the height of the launch point or a target to hit, and the laboratory finds the flight time, the maximum height, the range, the velocity at any instant, or the angle (often two) that reaches a given point.
\(x(t)\) is a straight line (uniform motion), \(y(t)\) a parabola (vertical throw), \(v_x\) constant and \(v_y\) decreasing at the rate \(g\).
The invariant is applied twice: along \(x\) with \(a_x = 0\) (equations 3 and 5) and along \(y\) with \(a_y = -g\) (4, 6 and 7), after splitting the launch velocity into components (1 and 2). The magnitude and direction of the velocity come from its components (8 and \(\varphi\)), and eliminating \(t\) gives the trajectory (T).
The landmarks of every flight: the top, the return to the launch level and the arrival at the ground.
In two dimensions the position is a vector \(\vec r = x\,\hat i + y\,\hat j\), and velocity and acceleration are its derivatives, \(\vec v = d\vec r/dt\) and \(\vec a = d\vec v/dt\). When the acceleration is constant, each component obeys the invariant on its own:
\[x = x_0 + v_{0x}t + \tfrac12 a_x t^2,\qquad y = y_0 + v_{0y}t + \tfrac12 a_y t^2.\]The velocity is always tangent to the path.
A projectile has no horizontal acceleration and a vertical acceleration \(-g\). So, as the course notes put it, the motion splits into a uniform motion along x with \(v_{0x} = v_0\cos\theta\) and a vertical throw along y with \(v_{0y} = v_0\sin\theta\). A ball shot horizontally and another one dropped at the same instant hit the ground together: the horizontal motion does not change the vertical fall.
From \(x = (v_0\cos\theta)\,t\) we get \(t = x/(v_0\cos\theta)\); substituting in \(y = (v_0\sin\theta)\,t - \tfrac12 g t^2\):
\[y = x\tan\theta - \frac{g\,x^2}{2\,(v_0\cos\theta)^2}.\]The highest power of \(y\) is 1 and that of \(x\) is 2: it is a parabola, and the angle \(\theta\) is its parameter. Because it does not contain \(t\), this equation is the one to use when the statement gives a point the projectile passes through.
The top is reached when \(v_y = v_0\sin\theta - gt = 0\), at \(t_s = \dfrac{v_0\sin\theta}{g}\), and there
\[Y_{\text{máx}} = \frac{v_0^2\sin^2\theta}{2g}.\]To come back to the launch level it needs twice that time, \(t_v = 2t_s\), and in that time it covers horizontally
\[X = (v_0\cos\theta)\,t_v = \frac{v_0^2\sin 2\theta}{g}.\]The range is largest at \(\theta = 45°\), and complementary angles (30° and 60°, say) give the same range.
If the range is \(k\) times the maximum height, \(\dfrac{v_0^2\sin 2\theta}{g} = k\,\dfrac{v_0^2\sin^2\theta}{2g}\). Writing \(\sin 2\theta = 2\sin\theta\cos\theta\) and cancelling \(\sin\theta\): \(4\cos\theta = k\sin\theta\), so
\[\tan\theta = \frac{4}{k}.\]For a range three times the maximum height, \(\tan\theta = 4/3\) and \(\theta = 53.1°\) (Serway 4.13); for a range equal to the height, \(\theta = 76.0°\).
To pass through a point \((x, y)\) with a given launch speed, write the trajectory with \(1/\cos^2\theta = 1 + \tan^2\theta\):
\[y = x\tan\theta - \frac{g x^2}{2v_0^2}\left(1 + \tan^2\theta\right),\]a quadratic in \(\tan\theta\). It usually has two roots, a flat trajectory and a high one; if the discriminant is negative the target is out of reach for that speed. With the angle given instead, the same equation gives the launch speed directly.
When the projectile leaves from a height \(h\) and lands on the ground, the end point is \(y = -h\) relative to the launch: the equations do not change, but the formulas \(t_v = 2t_s\) and \(X = v_0^2\sin 2\theta/g\) no longer hold. The flight time is the positive root of \(-h = (v_0\sin\theta)\,t - \tfrac12 g t^2\); as the course notes say, the negative root is discarded because there are no negative times.
If a gun aims straight at a target that is dropped at the instant of the shot, the bullet always hits it, whatever its speed. Without gravity the bullet would travel along the line of sight; gravity makes both the bullet and the target fall \(\tfrac12 g t^2\) below where they would have been, so they meet when the bullet reaches the vertical of the target (Resnick, section 4-3).
Problems adapted from Serway (chapters 2, 4, 5 and 6), Resnick and the course notes. Load any of them into the laboratory: the data fill the panel and the solver does the rest, so you can then change any value and see how the answer reacts.
A stone is thrown from the top of a building at an angle of \(30.0°\) with the horizontal and an initial speed of \(20.0\ \text{m/s}\). The building is \(45.0\ \text{m}\) high. (a) How long does the stone take to reach the ground? (b) What is its speed just before it hits the ground?
A soccer player kicks a ball at an angle of \(36°\) with the horizontal and an initial speed of \(15.5\ \text{m/s}\). Find (a) the time to reach the highest point, (b) its maximum height, (c) its range and time of flight, and (d) its velocity when it reaches the ground.
A cannon with a muzzle speed of \(1000\ \text{m/s}\) is used to destroy a target on top of a mountain, \(2000\ \text{m}\) away horizontally and \(800\ \text{m}\) above the ground. At what angle must it be fired? Ignore air resistance. (There are two answers: a flat shot and a high one.)
A cannon fires a ball at \(200\ \text{m/s}\) at an angle of \(40°\) with the horizontal. (a) Find the velocity and position of the ball after \(20\ \text{s}\). (b) Find the range and the time it takes to return to the ground (see the notable events).
A boy \(1.5\ \text{m}\) tall, standing \(15\ \text{m}\) from a wall \(5\ \text{m}\) high, throws a stone at \(45°\) with the horizontal. With what minimum speed must he throw it to clear the wall? (Measured from his hand, the top of the wall is at \(\Delta x = 15\ \text{m}\), \(\Delta y = 3.5\ \text{m}\).)
A plane flies horizontally at \(720\ \text{km/h}\) at a height of \(7840\ \text{m}\) and releases a bomb that explodes on reaching the ground. Calculate (a) the velocity of the bomb on reaching the ground and (b) the horizontal distance it travels.
A ball is thrown from a window on an upper floor of a building with an initial speed of \(8.00\ \text{m/s}\) at \(20.0°\) below the horizontal. It strikes the ground \(3.00\ \text{s}\) later. (a) How far horizontally from the base of the building does it strike? (b) From what height was it thrown?
A place kicker must kick a football from a point \(36.0\ \text{m}\) from the goal; the crossbar is \(3.05\ \text{m}\) high. The ball leaves the ground at \(20.0\ \text{m/s}\) at \(53.0°\). By how much does it clear or fall short of the crossbar? (Compare the height Δy at Δx = 36.0 m with 3.05 m.)
A physics student tries to jump a river on a motorcycle. The take-off ramp is inclined at \(53°\), the river is \(40.0\ \text{m}\) wide and the far bank is \(15.0\ \text{m}\) below the top of the ramp. What speed is needed at the top of the ramp to just reach the edge of the far bank? Then halve that speed and see where he lands.
In a contest to drop a package on a target, a plane flies horizontally at \(155\ \text{km/h}\) at a height of \(225\ \text{m}\). How long does the package fall and how far before the target must it be released? (The sight angle is the arctangent of that distance over the height.)
Theory after R. Resnick, D. Halliday and K. Krane, Física, vol. 1, chapter 4 (sections 4-1 to 4-3); problems from R. Serway and J. Jewett, Física para ciencias e ingeniería, vol. 1, chapter 4, and from the course notes on motion in the plane and the kinematics workshop.