Projectile motion
Specifying velocity by speed and angle or by components, deriving the equations of motion of a projectile from the acceleration vector, time of flight, greatest height, range, and the velocity at any time.
Setting up
A projectile launched with speed $V$ at angle $\theta$ above the horizontal has initial velocity $\begin{pmatrix} V\cos\theta \\ V\sin\theta \end{pmatrix}$. Ignoring air resistance, the only acceleration is gravity: $\underset{\sim}{a} = \begin{pmatrix} 0 \\ -g \end{pmatrix}$.
These come from integrating $\underset{\sim}{a}$ twice, as in the last section. In exams you are expected to derive them, not just quote them. The horizontal velocity never changes.
Key quantities
A ball is kicked at $20$ m/s at $30^\circ$ to the horizontal. Take $g = 10$ m/s$^2$. Find the time of flight, the greatest height and the range.
For the same ball, find its speed and direction at $t = 1.5$ s.
A ball is kicked at $20$ m/s at $30^\circ$. Take $g = 10$. Find the time of flight in seconds.
$10t - 5t^2 = 0$.
Find the greatest height of that ball, in metres.
$y(1) = 10 - 5$.
Find the range of that ball, in metres, correct to two decimal places.
$20\sqrt3$.
Find the speed of that ball at $t = 1.5$ s, correct to two decimal places.
$5\sqrt{13}$.
Find the horizontal component of that ball's velocity. Give an exact answer or a decimal.
$20\cos 30^\circ$.
A stone is thrown at $10$ m/s horizontally from the top of a cliff $45$ m high. Take $g = 10$. How far from the base of the cliff does it land, in metres?
$45 = 5t^2$ gives $t = 3$, so $x = 10 \times 3$.
For a projectile with no air resistance, the horizontal velocity:
There is no horizontal acceleration.
At the greatest height, the velocity is:
The vertical component is zero there.
The acceleration of a projectile is:
Only gravity acts.
New numbers every time. Keep going until you get five in a row.
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