Projectile from a Building: When and Where Does the Ball Land?

We model the motion of the ball as standard 2D projectile motion with an initial height.

---

1. Problem Setup

A ball is thrown from the top of a building:

• Building height: $h_0 = 50\,\text{m}$
• Initial speed: $v_0 = 20\,\text{m/s}$
• Launch angle: $\theta = 60^\circ$
• Gravitational acceleration: $g = 9.8\,\text{m/s}^2$ (downwards)

We assume no air resistance and take:

• Horizontal axis $x$: forward direction from the base of the building.
• Vertical axis $y$: height above the ground.

Initial position and velocity:

• $x(0) = 0$
• $y(0) = h_0 = 50$
• $v_{0x} = v_0 \cos\theta$
• $v_{0y} = v_0 \sin\theta$

For the given values:

$$\cos 60^\circ = \tfrac{1}{2},\quad \sin 60^\circ = \tfrac{\sqrt{3}}{2}$$

So:

$$\begin{aligned} v_{0x} &= 20 \cdot \tfrac{1}{2} = 10\,\text{m/s} \\ v_{0y} &= 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32\,\text{m/s} \end{aligned}$$

---

2. Equations of Motion

The parametric equations for projectile motion (with constant gravity) are:

$$\begin{aligned} x(t) &= v_{0x} t \\ y(t) &= h_0 + v_{0y} t - \tfrac{1}{2} g t^2 \end{aligned}$$

Substitute the values:

$$\begin{aligned} x(t) &= 10 t \\ y(t) &= 50 + 17.32 t - 4.9 t^2 \end{aligned}$$

We want to find when it hits the ground, i.e. when $y(t) = 0$.

---

3. Solve for Time of Impact

Set $y(t) = 0$:

$$50 + 17.32 t - 4.9 t^2 = 0$$

Rewriting in standard quadratic form:

$$-4.9 t^2 + 17.32 t + 50 = 0$$

Multiply by $-1$:

$$4.9 t^2 - 17.32 t - 50 = 0$$

Use the quadratic formula for $at^2 + bt + c = 0$:

$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here:

$$a = 4.9,\quad b = -17.32,\quad c = -50$$

So:

$$\begin{aligned} t &= \frac{-(-17.32) \pm \sqrt{(-17.32)^2 - 4(4.9)(-50)}}{2(4.9)} \\ &= \frac{17.32 \pm \sqrt{17.32^2 + 4\cdot4.9\cdot50}}{9.8} \end{aligned}$$

Compute the discriminant:

$$\begin{aligned} 17.32^2 &\approx 299.8 \\ 4\cdot4.9\cdot50 &= 980 \\ \Rightarrow \Delta &\approx 299.8 + 980 = 1279.8 \end{aligned}$$ $$\sqrt{1279.8} \approx 35.78$$

Therefore:

$$\begin{aligned} t_1 &= \frac{17.32 + 35.78}{9.8} \approx \frac{53.10}{9.8} \approx 5.42\,\text{s} \\ t_2 &= \frac{17.32 - 35.78}{9.8} \approx \frac{-18.46}{9.8} \approx -1.88\,\text{s} \end{aligned}$$

The negative time is not physically meaningful for our scenario, so the impact time is:

$$\boxed{t_{\text{hit}} \approx 5.42\,\text{s}}$$

---

4. Horizontal Distance at Impact

We now evaluate $x(t)$ at $t = t_{\text{hit}}$:

$$\begin{aligned} x_{\text{hit}} &= v_{0x} t_{\text{hit}} = 10 \cdot 5.42 \approx 54.2\,\text{m} \end{aligned}$$

So the ball lands about 54.2 meters horizontally from the point directly below the launch point (the base of the building).

---

5. Final Answer

• Time when the ball hits the ground: $\boxed{t \approx 5.42\,\text{s}}$
• Horizontal distance from the building: $\boxed{x \approx 54.2\,\text{m}}$

---

6. What the Visualization Shows

This interactive canvas illustrates:

1. Trajectory curve: The full parabolic path from the top of the building until just after impact.
2. Building: A vertical line at $x=0$ with a height of 50 m.
3. Dynamic projectile: A neon ball moving along the path, animated in real time.
4. Ground line: The horizontal axis (y = 0).
5. Impact marker: A highlight where the ball reaches the ground.

Widgets

You can adjust:

• The initial speed $v_0$
• The launch angle $\theta$
• The building height $h_0$
• The gravity $g$
• A time-scrubber (to freeze and inspect the motion at a specific time) or let it play automatically.

Mathematically, the visualization uses the general formulas:

$$\begin{aligned} v_{0x} &= v_0 \cos(\theta) \\ v_{0y} &= v_0 \sin(\theta) \\ x(t) &= v_{0x} t \\ y(t) &= h_0 + v_{0y} t - \tfrac12 g t^2 \end{aligned}$$

The impact time is computed each frame by solving $y(t) = 0$ using the quadratic formula, and the scale automatically adjusts to fit the entire motion (from the building top to the landing point) onto the canvas.