Projectiles example
If a girl on a beach kicks a ball into the sea at 7.2 ms-1 at an angle 胃 of 30掳 above the horizontal, the time of flight, maximum height reached and the range can all be worked out as follows.
Firstly the vertical and horizontal components of the initial velocity are calculated:
Vertical component equals:
\(v\sin (\theta)\)
\(= 7.2\times 0.5\)
\(= 3.6m{s^{ - 1}}\)
Horizontal component equals:
\(v\cos(\theta)\)
\(= 7.2\times0.866\)
\(=6.24m{s^{- 1}}\)
Time of flight
The time in the air is calculated using equations of motion as the vertical acceleration is Kept constant, or the same. Eg: the cars speed from A to B was uniform at 30mph. (uniform can be applied to other quantities such as force and acceleration).
\(u = 3.6m{s^{ - 1}}\) (upwards so positive)
\(v = 3.6m{s^{ - 1}}\) (negative as moving downwards at end of flight)
\(a = - 9.8m{s^{ - 2}}\) (negative as objects accelerate downwards due to gravity)
\(t = ?\)
Total time of flight, \({t_ {total}}\) using \(v = u + at\)
\(-3.6=3.6-9.8t\)
\(t = \frac{{ - 3.6 - 3.6}}{{ - 9.8}}\)
\(t = 0.735s\)
Maximum height
To calculate maximum height, we need to know the time taken to reach maximum height, \({t_{\max }}\).
The maximum height is reached when the vertical velocity reaches zero.
\(u = 3.6m{s^{ - 1}}\)
\(v = 0m{s^{ - 1}}\) (the ball is not moving in the vertical direction when at highest point)
\(a = - 9.8m{s^{ - 2}}\)
\(t = ?\)
\(t = \frac{{v - u}}{a} = \frac{{0 - 3.6}}{{ - 9.8}} = 0.367s\)
Maximum height reached, \({s_{\max }}\).
\(u = 3.6m{s^{ - 1}}\)
\(v = 0m{s^{ - 1}}\)
\(a = - 9.8m{s^{ - 2}}\)
\(s = ?\)
Any relevant equation of motion \(s = ut + \frac{1}{2}a{t^2}\), \({v^2} = {u^2} + 2as\), or \(s = \left({\frac{{u + v}}{2}}\right)t\) could be used.
\(s = ut + \frac{1}{2}a{t^2}\)
\(= 3.6 \times 0.367 + \frac{1}{2} \times ( - 9.8) \times {0.367^2}\)
\(= 0.661m\)
\(s = \frac{{{v^2} - {u^2}}}{{2a}}\)
\(= \frac{{{0^2} - {{3.6}^2}}}{{2 \times ( - 9.8)}}\)
\(= 0.661m\)
\(s = \left({\frac{{u + v}}{2}}\right)t\)
\(= \left( {\frac{{3.6 + 0}}{2}} \right) \times 0.367\)
\(= 0.661m\)
All three equations give the same answer of height = 0.66 metres above the sea.
Range
Finally the range is calculated using the horizontal component and time of flight.
\(v=6.24\,m{s^{- 1}}\)
\(t = 0.735s\)
\(d = ?\)
\(d = v \times t\)
\(d = 6.24 \times 0.735\)
\(= 4.6m\)