Paper 3 Projectile Motion Answers

Q1 Projectile landing below launch level

1. The standard formula assumes launch and landing are at the same vertical level. Here the package lands $25.0\mathrm{m}$ below the launch point, so the vertical time of flight must be found from component equations. [2]
2. $u_x=22.0\cos 28.0^{\circ }=19.4\mathrm{m}{\mathrm{s}}^{-1}$, $u_y=22.0\sin 28.0^{\circ }=10.3\mathrm{m}{\mathrm{s}}^{-1}$. Vertical displacement to ground is $y=-25.0\mathrm{m}$:

$$-25.0=10.3t-4.905t^2.$$

So $4.905t^2-10.3t-25.0=0$, giving $t=3.54\mathrm{s}$ for the positive root. [4] 3. Using unrounded values, $x=u_{x}t=68.8\mathrm{m}$. A value of $68.7\mathrm{m}$ is acceptable if rounded intermediate values are used consistently. [2] 4. $v_x=19.4\mathrm{m}{\mathrm{s}}^{-1}$. $v_y=u_y-gt=10.3-9.81(3.54)=-24.4\mathrm{m}{\mathrm{s}}^{-1}$. Hence

$$\vec{v}=(19.4\hat{\mathbf{i}}-24.4\hat{\mathbf{j}})\mathrm{m}{\mathrm{s}}^{-1}.$$

Speed $=31.2\mathrm{m}{\mathrm{s}}^{-1}$, direction $51.5^{\circ }$ below the horizontal. [4] 5. Air resistance would reduce the horizontal velocity during flight, so the range would usually be smaller than the ideal prediction. [2]

Q2 Trajectory equation

1. Horizontal motion gives $x=(u\cos \theta )t$, so $t=x/(u\cos \theta )$. Vertical motion gives $y=(u\sin \theta )t-\frac{1}{2}g{t}^2$. Substitute for $t$:

$$y=(u\sin \theta )\frac{x}{u\cos \theta }-\frac{1}{2}g{\left(\frac{x}{u\cos \theta }\right)}^2=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }.$$

[4] 2. Substitute $u=16.0\mathrm{m}{\mathrm{s}}^{-1}$, $\theta =40.0^{\circ }$, $x=12.0\mathrm{m}$:

$$y=12.0\tan 40.0^{\circ }-\frac{9.81(12.0{)}^2}{2(16.0{)}^2{\cos }^{2}40.0^{\circ }}=5.37\mathrm{m}.$$

[3] 3. The equation has the form $y=Ax-B{x}^2$, a quadratic in $x$. A quadratic displacement relation gives a parabolic path. [2] 4. Smaller $g$ gives smaller downward acceleration, longer time of flight and less downward curvature. For same-level launch, $R=u^2\sin 2\theta /g$, so smaller $g$ gives a larger range. [3]