Modelling a Fundraising Projectile Launch
A transition-year class is launching a model rocket for a charity event. The rocket is launched from a small platform 2 m above the ground. Its height h, in metres, above the ground after t seconds is modelled by h(t) = −4.9t² + 18t + 2, valid until the rocket returns to ground level. A banner is stretched horizontally at a height of 12 m, and a camera will be triggered only while the rocket is at or above the banner. The class needs the rocket to stay above the banner for at least 1.2 seconds to capture a usable video clip. The teacher also wants to know the maximum height reached and the total time the rocket is in the air, so the recovery team can position itself. The launch site is flat, and air resistance is ignored. One student claims that because the coefficient of t is 18, the rocket will be above 12 m for more than 2 seconds.
1. Solve the inequality h(t) ≥ 12 to find the time interval during which the rocket is at or above the banner. Give your answer correct to two decimal places.
Read the worked answer
Set
[
-4.9t^2+18t+2\ge 12.
]
Rearrange:
[
-4.9t^2+18t-10\ge 0.
]
Multiply by (-1) and reverse the inequality:
[
4.9t^2-18t+10\le 0.
]
Solve the corresponding equation:
[
t=\frac{18\pm\sqrt{18^2-4(4.9)(10)}}{2(4.9)}
=\frac{18\pm\sqrt{324-196}}{9.8}
=\frac{18\pm\sqrt{128}}{9.8}.
]
Since (\sqrt{128}\approx 11.31),
[
t_1\approx\frac{18-11.31}{9.8}\approx 0.68,
\qquad
t_2\approx\frac{18+11.31}{9.8}\approx 2.99.
]
The parabola opens downward, so (h(t)\ge 12) between the roots. Therefore the rocket is at or above the banner for
[
0.68\le t\le 2.99\text{ seconds}.
]
The duration is about (2.99-0.68=2.31) seconds, which is longer than the required 1.2 seconds.
2. Find the maximum height reached and the time at which it occurs. Does the rocket clear a 15 m safety line?
Read the worked answer
The height is a quadratic with (a=-4.9), (b=18), (c=2). The maximum occurs at
[
t=-\frac{b}{2a}=-\frac{18}{2(-4.9)}=\frac{18}{9.8}\approx 1.84\text{ s}.
]
Substitute:
[
h(1.84)=-4.9(1.84)^2+18(1.84)+2.
]
Now ((1.84)^2\approx 3.386), so
[
h\approx -4.9(3.386)+33.12+2
\approx -16.59+33.12+2\approx 18.53\text{ m}.
]
So the maximum height is about 18.53 m at about 1.84 s. Since (18.53>15), the rocket does clear the 15 m safety line.
3. Find the total time the rocket is in the air (to the nearest 0.01 s). Evaluate the student's claim that the rocket is above 12 m for more than 2 seconds, and comment on whether the coefficient 18 alone justifies the claim.
Read the worked answer
The rocket lands when (h(t)=0):
[
-4.9t^2+18t+2=0
\Rightarrow 4.9t^2-18t-2=0.
]
Using the quadratic formula:
[
t=\frac{18\pm\sqrt{(-18)^2-4(4.9)(-2)}}{2(4.9)}
=\frac{18\pm\sqrt{324+39.2}}{9.8}
=\frac{18\pm\sqrt{363.2}}{9.8}.
]
Since (\sqrt{363.2}\approx 19.06), the positive root is
[
t\approx\frac{18+19.06}{9.8}\approx 3.78\text{ s}.
]
So the total flight time is about 3.78 seconds.
From the first task, the rocket is above 12 m from (t\approx0.68) s to (t\approx2.99) s, a duration of about (2.31) s. The student's claim that it is above 12 m for more than 2 seconds is therefore correct.
However, the coefficient 18 alone does not justify the claim. The constant term 2 raises the launch height, and the (-4.9t^2) term controls the downward acceleration. Changing the constant or the quadratic coefficient would change the duration above 12 m even if the coefficient of (t) stayed 18. The claim is true only because of the full calculation, not because of the size of the linear coefficient alone.