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Module 5: Advanced Mechanics
Quick questions on Projectile motion explained: HSC Physics Module 5
9short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.
What is resolving the initial velocity?Show answer
If a projectile is launched with speed $v_0$ at angle $\theta$ above the horizontal:
What is horizontal motion?Show answer
No horizontal force acts (air resistance is ignored), so horizontal velocity is constant.
What is vertical motion?Show answer
The only acceleration is gravity, $a_y = -g$ (taking up as positive). Use SUVAT:
What is key features of the trajectory?Show answer
The path is a parabola. At maximum height $v_y = 0$, so $h_{\max} = \frac{v_{0y}^2}{2g}$.
What is mixing up horizontal and vertical equations?Show answer
Horizontal velocity never changes (in HSC, where we ignore air resistance). Vertical velocity changes by $-9.8 \text{ m/s}$ every second. Set up two separate columns of working.
What is forgetting the sign of $g$?Show answer
If you take up as positive, $g$ enters the SUVAT equations as $-9.8 \text{ m/s}^2$. If you take down as positive, $g$ is $+9.8 \text{ m/s}^2$. Pick a convention and stick to it for the whole question.
What is using the speed instead of a component?Show answer
$v_0 = 25$ m/s at $40°$ does not mean the horizontal velocity is $25$ m/s. You must resolve into components first.
What is treating horizontally thrown objects as having $v_{0y} = v_0$?Show answer
If a stone is thrown horizontally off a cliff, $v_{0y} = 0$. The full speed is in the horizontal direction.
What is forgetting units?Show answer
Markers deduct for missing units (m, s, m/s) even when the number is correct.