Quadratic Word Problems: Projectile Motion

Setting up the problem

A ball is thrown upward from height 1 m with initial velocity 20 m/s. Its height at time $t$ is:

$h(t) = -5t^2 + 20t + 1$

We need to find: (a) when it hits the ground, (b) the maximum height, (c) total flight time.

Part (a): When does it hit the ground?

The ball hits the ground when $h(t) = 0$:

$-5t^2 + 20t + 1 = 0$

Step 1 — Apply the quadratic formula with $a = -5$, $b = 20$, $c = 1$:

$t = \frac{-20 \pm \sqrt{400 + 20}}{-10} = \frac{-20 \pm \sqrt{420}}{-10}$

Step 2 — Compute: $\sqrt{420} \approx 20.49$

$t_1 = \frac{-20 + 20.49}{-10} = \frac{0.49}{-10} \approx -0.05 \text{ s (rejected — before throw)}$

$t_2 = \frac{-20 - 20.49}{-10} = \frac{-40.49}{-10} \approx 4.05 \text{ s}$

Answer: The ball hits the ground at $t \approx 4.05$ seconds.

Part (b): Maximum height

The vertex of the parabola gives the maximum height. The time of maximum height is:

$t_{\text{max}} = \frac{-b}{2a} = \frac{-20}{2(-5)} = \frac{-20}{-10} = 2 \text{ s}$

Plug back in:

$h(2) = -5(4) + 20(2) + 1 = -20 + 40 + 1 = 21 \text{ m}$

Answer: Maximum height is 21 meters at $t = 2$ seconds.

Part (c): Total flight time

From launch ($t = 0$) to landing ($t \approx 4.05$ s), the total flight time is about 4.05 seconds.

Key formulas for projectile problems

• Height function: $h(t) = -\frac{1}{2}gt^2 + v_0 t + h_0$
• Time to max height: $t = v_0/g$
• Max height: plug $t$ back into $h(t)$
• Landing time: solve $h(t) = 0$ using the quadratic formula

Try it in the visualization

Adjust the initial velocity and height. The parabola shows the trajectory over time. Maximum height and landing time are marked. The ball animates along the curve.