Paper 2 Projectile Motion Answers

Q1 Horizontal launch

1. $9.81\mathrm{m}{\mathrm{s}}^{-2}$ downward. [1]
2. Vertical motion: $0.800=\frac{1}{2}g{t}^2$, so $t=\sqrt{2(0.800)/9.81}=0.404\mathrm{s}$. [2]
3. $x=v_{x}t=2.40(0.404)=0.970\mathrm{m}$. [2]
4. $v_x=2.40\mathrm{m}{\mathrm{s}}^{-1}$, $v_y=gt=3.96\mathrm{m}{\mathrm{s}}^{-1}$ downward. Speed $=\sqrt{2.40^2+3.96^2}=4.63\mathrm{m}{\mathrm{s}}^{-1}$. Angle below horizontal $={\tan }^{-1}(3.96/2.40)=58.8^{\circ }$. [4]

Q2 Launch at an angle

1. $u_x=18.0\cos 35.0^{\circ }=14.7\mathrm{m}{\mathrm{s}}^{-1}$, $u_y=18.0\sin 35.0^{\circ }=10.3\mathrm{m}{\mathrm{s}}^{-1}$. [2]
2. At highest point $v_y=0$, so $t=u_y/g=10.3/9.81=1.05\mathrm{s}$. [2]
3. $h=u_y^2/(2g)=10.3^2/(2\times 9.81)=5.43\mathrm{m}$. [2]
4. Total flight time $=2u_y/g=2.10\mathrm{s}$. Range $=u_{x}t=14.7(2.10)=31.0\mathrm{m}$. [3]
5. The vertical velocity is zero only instantaneously. Gravity still acts downward, so the resultant force and acceleration are downward. [2]

Q3 Component reasoning

1. Air resistance is negligible, or both balls have the same vertical acceleration $g$. [1]
2. Both balls have the same initial vertical velocity, zero. Both have the same vertical displacement to the floor and the same vertical acceleration. The time of fall is determined by vertical motion only, not by horizontal velocity. [3]
3. For example, ball B landing later than ball A, or a lighter/less streamlined ball deviating from the predicted parabolic path/range, would indicate air resistance is significant. [2]