A small rocket is fired into the air from a fixed position on the ground - Leaving Cert Mathematics - Question 6 - 2014

Draw the graph using the completed table values: plot the points (t, h) for each time and connect them smoothly to show the trajectory of the rocket. The peak should occur around $(5, 25)$.

From the graph, estimate the height at $t = 2.5$ seconds, which is approximately $19$ m.

The corresponding time for the rocket returning to this height of $19$ m appears to be around $7.5$ seconds.

The highest point reached by the rocket is at the coordinate $(5, 25)$.

To find the slope, use the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{21 - 24}{7 - 6} = -3$

Yes. The slope of the line joining (7, 21) and (8, 16) is given by:

$m = \frac{16-21}{8-7} = -5$ The line is steeper because the absolute value of the slope is greater.

Differentiating $h = 10t - t^2$, we find:

$\frac{dh}{dt} = 10 - 2t$

To find the maximum height, set the derivative to zero:

$10 - 2t = 0 \implies t = 5\ seconds$ Then substitute $t = 5$ back into the height formula:

$h = 10(5) - (5)^2 = 25\ m$

Substituting $t = 3$ into the derivative:

$S = \frac{dh}{dt} = 10 - 2(3) = 4\ m/s$

Set the derivative to 2:

$10 - 2t = 2 \implies t = 4\ seconds$ Then substitute into the height formula:

$h = 10(4) - (4)^2 = 24\ m$ Thus, the coordinates are $(4, 24)$.