Motion of a projectile — practice questions

First, resolve the initial velocity into components. Let upwards and rightwards be positive. $u_x = 42 \cos 30^\circ = 42 \times \frac{\sqrt{3}}{2} = 21\sqrt{3} \, \text{m s}^{-1}$ $u_y = 42 \sin 30^\circ = 42 \times \frac{1}{2} = 21 \, \text{m s}^{-1}$

(a) Greatest Height At the greatest height, the vertical velocity $v_y = 0$. We analyse the vertical motion: $s_y = ?$, $u_y = 21$, $v_y = 0$, $a_y = -9.8$ Using $v^2 = u^2 + 2as$: $0^2 = 21^2 + 2(-9.8)s_y$ $19.6 s_y = 441$ $s_y = \frac{441}{19.6} = 22.5 \, \text{m}$ So, the greatest height reached is 22.5 m. [M1 for using $v_y=0$, A1 for correct substitution, A1 for answer]

(b) Time of Flight The ball lands when its vertical displacement $s_y = 0$. We analyse the vertical motion: $s_y = 0$, $u_y = 21$, $a_y = -9.8$, $t = ?$ Using $s = ut + \frac{1}{2}at^2$: $0 = 21t + \frac{1}{2}(-9.8)t^2$ $0 = 21t - 4.9t^2$ $0 = t(21 - 4.9t)$ This gives $t=0$ (the start) or $21 - 4.9t = 0$. $t = \frac{21}{4.9} \approx 4.2857... \, \text{s}$ The time of flight is $4.29 \, \text{s}$ (3 s.f.). [M1 for setting $s_y=0$, A1 for correct quadratic, A1 for answer]

(c) Range The range is the horizontal distance travelled during the time of flight. We use the horizontal motion: $s_x = ?$, $u_x = 21\sqrt{3}$, $t = 4.2857...$ $s_x = u_x t$ $s_x = (21\sqrt{3}) \times (4.2857...)$ $s_x = 155.88... \, \text{m}$ The range is $156 \, \text{m}$ (3 s.f.). [M1 for using Range = $u_x \times t_{flight}$, A1 for answer]

Let's set up a coordinate system with the x-axis parallel to the plane (upwards) and the y-axis perpendicular to the plane (outwards).

Resolve initial velocity: $u_x = 30 \cos 45^\circ = 15\sqrt{2} \, \text{m s}^{-1}$ $u_y = 30 \sin 45^\circ = 15\sqrt{2} \, \text{m s}^{-1}$

Resolve acceleration due to gravity: $a_x = -g \sin 20^\circ = -9.8 \sin 20^\circ \approx -3.3518 \, \text{m s}^{-2}$ $a_y = -g \cos 20^\circ = -9.8 \cos 20^\circ \approx -9.2088 \, \text{m s}^{-2}$

(a) Time of Flight The particle lands on the plane when its displacement perpendicular to the plane is zero ($s_y = 0$). Using $s_y = u_y t + \frac{1}{2}a_y t^2$: $0 = (15\sqrt{2})t + \frac{1}{2}(-9.8 \cos 20^\circ)t^2$ $0 = t(15\sqrt{2} - 4.9 \cos 20^\circ \, t)$ $t=0$ or $t = \frac{15\sqrt{2}}{4.9 \cos 20^\circ} \approx \frac{21.213}{4.6044} \approx 4.607... \, \text{s}$ The time of flight is $4.61 \, \text{s}$ (3 s.f.). [M1 for resolving g, M1 for setting $s_\perp=0$, A1 for correct equation, A1 for answer]

(b) Range along the plane The range is the displacement parallel to the plane ($s_x$) at the time of flight. Using $s_x = u_x t + \frac{1}{2}a_x t^2$ with $t = 4.607... \, \text{s}$: $s_x = (15\sqrt{2})(4.607...) + \frac{1}{2}(-9.8 \sin 20^\circ)(4.607...)^2$ $s_x = (21.213)(4.607...) + \frac{1}{2}(-3.3518)(21.228...)$ $s_x = 97.735... - 35.569...$ $s_x = 62.166... \, \text{m}$ The range along the plane is $62.2 \, \text{m}$ (3 s.f.). [M1 for using $s_\parallel = u_\parallel t + \frac{1}{2} a_\parallel t^2$ with their time of flight, A1 for answer]