projectile motion in a uniform gravitational field

Projectile motion in a uniform gravitational field is motion with initial velocity $\vec v_0$ in which the only acceleration is $\vec g$, directed vertically downward. With the $x$-axis horizontal and the $y$-axis vertically upward, the acceleration components are $a_x = 0$ and $a_y = -g$, so the motion splits into two independent one-dimensional motions. Because $\vec v = \vec v_0 + \vec g t$ and $\vec r = \vec r_0 + \vec v_0 t + \frac{1}{2}\vec g t^2$, the horizontal velocity $v_x = v_0\cos\alpha$ stays constant, while the vertical velocity $v_y = v_0\sin\alpha - gt$ changes uniformly. The corresponding coordinates are $x = x_0 + v_0\cos\alpha\, t$ and $y = y_0 + v_0\sin\alpha\, t - \frac{1}{2}gt^2$. Eliminating $t$ gives the trajectory equation $y = y_0 + (x-x_0)\tan\alpha - \frac{g(x-x_0)^2}{2v_0^2\cos^2\alpha},$ which is a parabola.

A ball is thrown from ground level with speed $v_0 = 20\ \text{m/s}$ at an angle $\alpha = 30^\circ$ to the horizontal. Taking $g = 10\ \text{m/s}^2$ and neglecting air resistance, find the time of flight and the horizontal range.

A projectile is launched with speed $v_0 = 15\ \text{m/s}$ at an angle $\alpha = 40^\circ$ to the horizontal from a cliff of height $h = 20\ \text{m}$. Taking $g = 9.8\ \text{m/s}^2$, find the time until it hits the ground and the horizontal distance from the launch point.

A projectile is launched from level ground with a fixed initial speed $v_0$. (a) Derive the range formula $R(\alpha)$ and show that the range is maximal at $\alpha = 45^\circ$. (b) Find the two launch angles for which the range equals $0.8R_{\max}$.

Use a projectile-motion simulation (for example, PhET "Projectile Motion"); do not perform measurements with real thrown objects. Set $v_0 = 20\ \text{m/s}$ and launch height $0$. Vary the angle from $15^\circ$ to $75^\circ$ in steps of $10^\circ$, recording the range in a table. Plot $R(\alpha)$ and mark the maximum. Verify that complementary angles adding to $90^\circ$ (for example, $30^\circ$ and $60^\circ$) give the same range. Compare your data with $R = \frac{v_0^2\sin 2\alpha}{g}$, then switch on air resistance and describe which conclusions no longer hold.

Show that projectile motion is the superposition of uniform horizontal motion and uniformly accelerated vertical motion, and that this superposition produces a parabolic trajectory.

A sketch of the parabolic trajectory on $x$–$y$ axes: the initial velocity vector $\vec v_0$ with its components $v_{0x}$, $v_{0y}$; velocity vectors at several points of the trajectory, whose horizontal component stays unchanged while the vertical component decreases, vanishes at the apex, and then reverses; vertical dashed lines marking equal time intervals and vertical drops growing as $t^2$.

At the free-throw line, a player releases the ball from a height of $2.0\ \text{m}$ with speed $v_0=7.5\ \text{m/s}$ at $\theta=50^\circ$ above the horizontal. The hoop is $4.6\ \text{m}$ away horizontally and $3.05\ \text{m}$ high. The horizontal velocity is $v_x=v_0\cos\theta=4.82\ \text{m/s}$, so the ball reaches the hoop's horizontal position after $t=x/v_x=0.954\ \text{s}$. Its height then is $y=2.0+v_0\sin\theta\,t-\frac12 g t^2=3.02\ \text{m}$, just below the rim.

The example uses the independence of horizontal and vertical motion: $x(t)=v_x t$ with constant $v_x$ gives the time, while $y(t)=y_0+v_{0y}t-\frac12 g t^2$ with $a_y=-g$ gives the height. The trajectory is the parabola formed by combining these two equations.

A stunt car leaves a horizontal ramp with speed $v_0=20\ \text{m/s}$ from a height $h=5.0\ \text{m}$ above the landing zone. With zero initial vertical velocity, the fall time is $t=\sqrt{2h/g}=1.01\ \text{s}$. In that time it travels horizontally $x=v_0t=20.2\ \text{m}$. Its vertical speed at landing is $v_y=gt=9.9\ \text{m/s}$, so its total speed is $\sqrt{v_0^2+v_y^2}=22.3\ \text{m/s}$.

This scenario exercises projectile motion with a horizontal launch: $v_{0y}=0$, but $a_y=-g$ still acts. The horizontal displacement comes from constant $v_x$, and the vertical velocity comes from uniformly accelerated motion; the final velocity is the vector sum.

While walking at $1.4\ \text{m/s}$, a student accidentally lets a phone slip from a height of $1.5\ \text{m}$. The phone starts with horizontal velocity $1.4\ \text{m/s}$. It falls for $t=\sqrt{2h/g}=0.553\ \text{s}$, lands $x=1.4t=0.77\ \text{m}$ ahead of the release point, and has vertical impact speed $v_y=gt=5.42\ \text{m/s}$.

The phone is a projectile because the only significant force after release is gravity. The horizontal component $v_x=1.4\ \text{m/s}$ stays constant, while the vertical component grows as $v_y=gt$, exactly as the component equations predict.