Solution |

Using a graph, we don’t need to solve the simultaneous
equations algebraically. Instead, we can just graph the two equations on the
same set of axes. Both equations are written in the form, “height =
something…” This means that where the two lines Sponsored Links The linear equation is easy to plot, it’s just a straight line starting at the origin and sloping upwards to the right. The parabola representing the bullet’s height is a bit more complicated. First we need to write the equation in standard form: Now, instead of ‘x’s, we have ‘t’s, so let’s rewrite this standard quadratic form: Our equation for the bullet’s height is So this means a = 5, b = –20 and c = h Notice how we only have a So the coordinates of the turning point in the graph are (2 s, 20 m). The other thing we need to do is work out our ‘y’ axis intercept – in this case it’s our ‘h’ axis intercept. This occurs when our ‘x’ value is equal to ‘0’ – in this question when our ‘t’ value is equal to ‘0’: So our ‘h’ axis intercept is at 0 m. This makes sense –
at time 0, The part of the parabola below the t-axis doesn’t really need to be there – it indicates that after 4 seconds the bullet will have fallen past Jason and be going below him. Most likely in real life it will have hit the ground and stopped. I’ve continued the parabola below the t-axis just to give a better overall idea of its shape. Don’t be tricked by the shape of the parabola – it So back to our first graph, showing the vertical height of the bullet plotted as a function of time. We must not forget to plot the line showing the balloon’s height above the ground as well. This one’s easy to plot since it’s a straight line, so we can quickly add it to the graph: Now, when the lines touch and cross over each other, we
know at that time the bullet and the balloon are at exactly the same height.
This happens at two times in the graph – at time t = 0, and at around time t
= 3.6 seconds. Let’s drop a vertical line from the 2 It looks like the cross-over point is at around t = 3.6 seconds. We can also read off the height of the bullet and the balloon at that time by reading the value of ‘h’ at that time. From the graph it looks like h is around 7.5 metres. Now is the bullet travelling upwards or downwards at this
time? Well, the parabola is sloping Comparing our graph answers with our algebraic answers, our time answers are exactly the same, but our height answers are a little bit different, caused by the inexactness of the graph. |