At a school sports day, the height of a thrown shot above the ground, \(h\) metres, \(t\) seconds after release, is modelled by \(h = -5t^2 + 8t + 1.8\), sh...
Question 1 Report
At a school sports day, the height of a thrown shot above the ground, \(h\) metres, \(t\) seconds after release, is modelled by \(h = -5t^2 + 8t + 1.8\), shown below.
- Write down the height at which the shot is released. (1)
- Use the quadratic formula to find the time at which the shot lands, rejecting any solution that does not fit the context. (3)
- By writing \(h\) in completed-square form, find the maximum height of the shot. (2)
- A nearby high-jump crossbar is fixed at \(5.2\) m. Using your answer to part (c), state, with a reason, whether the shot ever reaches this height. (1)
Answer Details
This question reads an initial value from a projectile model, uses the quadratic formula to find when it lands, and uses completed-square form to find and check its maximum height against a fixed limit.
At release, \(t=0\), so:
\[h = -5(0)^2+8(0)+1.8 = 1.8 \text{ m}\][1 mark]
The shot lands when \(h=0\):
\[-5t^2+8t+1.8=0\]Multiplying by \(-10\) to clear the decimal and make the leading coefficient positive:
\[50t^2-80t-18=0 \Rightarrow 25t^2-40t-9=0\][1 mark]
Using the quadratic formula with \(a=25\), \(b=-40\), \(c=-9\):
\[t = \dfrac{40 \pm \sqrt{(-40)^2-4(25)(-9)}}{2(25)} = \dfrac{40 \pm \sqrt{2500}}{50} = \dfrac{40\pm50}{50}\]giving \(t=1.8\) or \(t=-0.2\) [1 mark]. Time cannot be negative, so \(t=1.8\) seconds [1 mark].
Completing the square:
\[h = -5(t^2-1.6t)+1.8 = -5\left[(t-0.8)^2-0.64\right]+1.8 = -5(t-0.8)^2+5\]The maximum height is \(5\) m, occurring at \(t=0.8\) s [2 marks].
Since the maximum height reached, \(5\) m, is less than \(5.2\) m, the shot never reaches the height of the crossbar [1 mark].
Because part (c)'s completed-square form gives an exact maximum of \(5\) m, no further checking of intermediate values of \(t\) is needed to answer part (d): the entire path never exceeds \(5\) m.
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