Quadratics: Question 8

Worked solution

Part (a): Factorising to find when the ball is at ground level

$h(t) = -5t^2 + 30t = -5t(t - 6)$

Setting $h(t)=0$:

$-5t(t-6) = 0 \implies t = 0 \quad \text{or} \quad t = 6$

Check by expanding back: $-5t(t-6) = -5t^2+30t$, which matches $h(t)$. ✓

In context: $t=0$ s is the instant the ball is thrown (height $0$ at launch), and $t=6$ s is the instant the ball lands back on the ground.

Part (b): Does the ball ever reach 50 m?

Set $h(t)=50$:

$-5t^2 + 30t = 50$

$-5t^2 + 30t - 50 = 0$

Divide through by $-5$:

$t^2 - 6t + 10 = 0$

Here $a=1$, $b=-6$, $c=10$, so the discriminant is:

$b^2-4ac = (-6)^2 - 4(1)(10) = 36 - 40 = -4$

Since the discriminant is negative, this equation has no real solutions. There is no real time $t$ at which the height equals $50$ m. So the ball never reaches a height of $50$ metres.

(This makes sense: the maximum height, found by completing the square on $h(t)=-5(t-3)^2+45$, is $45$ m, which is below $50$ m.)

Part (c): Solving exactly for a height of 35 m

Set $h(t)=35$:

$-5t^2 + 30t = 35$

$-5t^2 + 30t - 35 = 0$

Divide through by $-5$:

$t^2 - 6t + 7 = 0$

Apply the quadratic formula with $a=1$, $b=-6$, $c=7$:

$\text{discriminant} = (-6)^2 - 4(1)(7) = 36 - 28 = 8$

$t = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$

Numerical check: $\sqrt{2}\approx1.414$, so $t\approx1.586$ or $t\approx4.414$, both lie between $t=0$ and $t=6$, as expected for a height reached once on the way up and once on the way down.

Check: at $t=3-\sqrt2$, $h(t) = -5(3-\sqrt2)^2+30(3-\sqrt2) = -5(9-6\sqrt2+2)+90-30\sqrt2 = -5(11-6\sqrt2)+90-30\sqrt2 = -55+30\sqrt2+90-30\sqrt2 = 35$. ✓

Final answers

• (a) $t = \boxed{0}$ s (launch) and $t = \boxed{6}$ s (landing)
• (b) The ball never reaches 50 m, discriminant $=\boxed{-4}<0$
• (c) $t = \boxed{3 - \sqrt{2}}$ s or $t = \boxed{3 + \sqrt{2}}$ s