A ball is thrown upward at 10 m/s from the top of a 15 m tower. Find (a) the time for the ball to reach the ground and (b) the maximum height above the ground.

What this motion looks like

The ball starts 15 m above the ground, moves upward at first, slows down under gravity, then turns around and falls past the launch point to the ground. We can model the vertical position with a single quadratic equation.

Writing the height function

Take upward as positive and measure height from the ground.

• Initial height: $y_0=15\,\text{m}$
• Initial velocity: $u=10\,\text{m/s}$
• Acceleration: $a=-g=-9.8\,\text{m/s}^2$

So, $$y(t)=y_0+ut-\tfrac{1}{2}gt^2=15+10t-4.9t^2.$$

(a) Time to reach the ground

The ball reaches the ground when $y(t)=0$: $$15+10t-4.9t^2=0$$ Rewrite: $$4.9t^2-10t-15=0.$$ Quadratic formula: $$t=\frac{10\pm\sqrt{(-10)^2-4(4.9)(-15)}}{2(4.9)} =\frac{10\pm\sqrt{100+294}}{9.8} =\frac{10\pm\sqrt{394}}{9.8}.$$ The physical (positive) root is $$t=\frac{10+\sqrt{394}}{9.8}\approx 3.05\,\text{s}.$$ (The negative root corresponds to a time before the throw.)

(b) Maximum height above the ground

At the top point, the velocity is zero: $$v(t)=u-gt=0 \quad\Rightarrow\quad t_{\text{top}}=\frac{u}{g}=\frac{10}{9.8}\approx 1.02\,\text{s}.$$ The rise above the launch point is $$\Delta h=\frac{u^2}{2g}=\frac{10^2}{2(9.8)}=\frac{100}{19.6}\approx 5.10\,\text{m}.$$ So the maximum height above ground is $$h_{\max}=15+\Delta h\approx 15+5.10=20.1\,\text{m}.$$

Quick check (sanity)

Since it starts at 15 m and goes up about 5 m, a peak near 20 m makes sense. A total flight time a bit over 3 s is also reasonable because it takes about 1 s to go up, then longer to fall from about 20 m down to 0 m.