Paper 1 Projectile Motion Answers

Q	Ans	Brief reason
1	B	In the ideal model, $a_x=0$, so $v_x$ is constant.
2	B	For an angle measured above the horizontal, the vertical component is $u\sin \theta$.
3	B	$v_y=0$ at the apex, but $a_y=-g$.
4	B	Gravity acts vertically only.
5	B	$x=v_{x}t=3.0(0.40)=1.2\mathrm{m}$.
6	B	Same-level flight time is $2u_y/g$.
7	B	$\sin 2\theta$ is maximum when $2\theta =90^{\circ }$.
8	A	$x=u_{x}t$ for constant $v_x$.
9	B	$y=u_{y}t-\frac{1}{2}g{t}^2$.
10	C	$v=\sqrt{6^2+8^2}=10\mathrm{m}{\mathrm{s}}^{-1}$.
11	B	Acceleration remains the gravitational acceleration.
12	A	It retains the train’s horizontal velocity in the ground frame.
13	B	$u_x=20\cos 30^{\circ }=17.3\mathrm{m}{\mathrm{s}}^{-1}$.
14	B	The derivation assumes same launch and landing height.
15	B	Vertical motion controls time of fall from a given height.
16	C	The same elapsed time applies to both components.
17	B	Landing below launch point means negative vertical displacement.
18	B	Velocity is built from velocity components.
19	C	The vertical component reverses sign for same-level landing.
20	D	Horizontal acceleration is zero in the ideal model.