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

To draw the graph, plot the points from the completed table on a grid, with the x-axis representing time (t) and the y-axis representing height (h). Connect the points with a smooth curve to reflect the trajectory of the rocket.

From the graph, at t = 2.5 seconds, the height of the rocket is approximately 19 meters.

The rocket will again be at this height of 19 meters at approximately 7.5 seconds.

The highest point reached by the rocket is at the coordinates (5, 25), where 5 seconds is the time and 25 meters is the height.

The slope (m) of the line joining the points (6, 24) and (7, 21) is calculated using the formula:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{21 - 24}{7 - 6} = \frac{-3}{1} = -3.$

Yes, the line joining the points (7, 21) and (8, 16) is expected to be steeper because the absolute value of the slope is greater for this segment, calculated as:
$m = \frac{16 - 21}{8 - 7} = \frac{-5}{1} = -5.$

To find the derivative of height with respect to time, we differentiate the height equation:
$\frac{dh}{dt} = 10 - 2t.$

To find the maximum height, set rac{dh}{dt} = 0:
$10 - 2t = 0 \Rightarrow t = 5.$
Substituting $t = 5$ into the height equation gives:
h = 10(5) - (5)^2 = 25 m.$$

The speed of the rocket at t = 3 seconds is given by:
$S = \frac{dh}{dt} = 10 - 2(3) = 4 \text{ m/s}.$

To find the point where the slope is 2, we set the derivative equal to 2:
$10 - 2t = 2 \Rightarrow 2t = 8 \Rightarrow t = 4.$
Now, substituting $t = 4$ into the height equation gives:
h = 10(4) - (4)^2 = 24.
Hence, the coordinates are (4, 24).