Worked exercise — Range of a projectile

Problem

A projectile leaves at $40\;\text{m/s}$ with an angle of $45^\circ$. Calculate the range and the maximum height (neglecting air resistance).

Solution

Initial components. Let us resolve the initial speed along the axes. With $\theta = 45^\circ$ and $v_0 = 40\;\text{m/s}$:

$v_{0x} = v_0\cos 45^\circ = \frac{40}{\sqrt 2} \approx 28{,}28\;\mathrm{m/s}, \qquad v_{0y} = v_0\sin 45^\circ \approx 28{,}28\;\mathrm{m/s}$

As with free fall in 2D, the motion is the superposition of a uniform horizontal motion (speed $v_{0x}$) and a uniformly accelerated vertical motion with acceleration $-g$.

Range. This is the horizontal distance covered when the projectile returns to the launch height. Using the general formula:

$x_\text{max} = \frac{v_0^2\sin 2\theta}{g} = \frac{40^2 \cdot \sin 90^\circ}{9{,}81} = \frac{1600 \cdot 1}{9{,}81} \approx 163\;\mathrm{m}$

Maximum height. At the highest point the vertical component of the speed vanishes ($v_y = 0$). Using $v_y^2 = v_{0y}^2 - 2g\,h$ with $v_y = 0$:

$h_\text{max} = \frac{v_0^2\sin^2\theta}{2g} = \frac{1600 \cdot 0{,}5}{19{,}62} \approx 40{,}8\;\mathrm{m}$

At $45^\circ$ the maximum range is obtained for a given $v_0$ (without friction): it is the angle that balances horizontal speed and flight time.

$\ev{x_\text{max} \approx 163\;\mathrm{m}, \quad h_\text{max} \approx 40{,}8\;\mathrm{m}}$

Links

Topics: Cinematica · Vettori Concepts: Moto parabolico Skills: Scomposizione vettoriale · Uso della trigonometria