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

We model the ball as an ideal projectile launched from a height of 50 m with initial speed 20 m/s at an angle of 60° above the horizontal, under constant gravitational acceleration.

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1. Set up the equations of motion

Let:

• Initial height: $y_0 = 50\,\text{m}$
• Initial speed: $v_0 = 20\,\text{m/s}$
• Launch angle: $\theta = 60^\circ$
• Gravitational acceleration: $g = 9.8\,\text{m/s}^2$ (downwards)

Horizontal and vertical components of the initial velocity:

$$v_{0x} = v_0 \cos\theta, \quad v_{0y} = v_0 \sin\theta.$$

For $\theta = 60^\circ$:

$$\cos 60^\circ = 0.5, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866.$$

So:

$$v_{0x} = 20 \cdot 0.5 = 10\,\text{m/s},$$ $$v_{0y} = 20 \cdot 0.866 \approx 17.32\,\text{m/s}.$$

Assuming the ground is at $y = 0$ and the ball is thrown from $(x, y) = (0, 50)$, the motion is:

• Horizontal:

$$x(t) = v_{0x} t = 10 t.$$

• Vertical:

$$y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2.$$

With values:

$$y(t) = 50 + 17.32 t - 4.9 t^2.$$

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2. Find the time when it hits the ground

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

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

Rewrite as a standard quadratic:

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

Multiply by $-1$ to make the leading coefficient positive:

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

Use the quadratic formula for $a = 4.9$, $b = -17.32$, $c = -50$:

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

So:

$$t = \frac{17.32 \pm \sqrt{(-17.32)^2 - 4 \cdot 4.9 \cdot (-50)}}{2 \cdot 4.9}.$$

Compute the discriminant:

$$(-17.32)^2 \approx 299.83, \quad 4 \cdot 4.9 \cdot (-50) = -980.$$

So:

$$\Delta = 299.83 - (-980) = 299.83 + 980 \approx 1279.83.$$ $$\sqrt{\Delta} \approx \sqrt{1279.83} \approx 35.78.$$

Thus:

$$t = \frac{17.32 \pm 35.78}{9.8}.$$

Two solutions:

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

The negative time is not physically meaningful (it would correspond to a time before the throw), so the impact time is:

$$\boxed{t_\text{hit} \approx 5.4\,\text{s}}.$$

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3. Find the horizontal distance at impact

Use $x(t) = v_{0x} t$ with $v_{0x} = 10\,\text{m/s}$ and $t_\text{hit} \approx 5.42\,\text{s}$:

$$x_\text{hit} = 10 \cdot 5.42 \approx 54.2\,\text{m}.$$

So the ball lands about 54 meters horizontally from the base of the building.

$$\boxed{x_\text{hit} \approx 54\,\text{m}}.$$

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4. What this visualization shows

This interactive canvas lets you:

• Vary the initial speed $v_0$ and angle $\theta$.
• Adjust the launch height (default 50 m) and gravity (default ~9.8 m/s²).
• See the trajectory curve of the ball from the building down to the ground.
• Watch a moving ball sliding along the physically correct path.
• Observe the predicted impact point on the ground, and how it changes as you move sliders.

Mathematically, at any time $t$ the position is:

$$\begin{aligned} x(t) &= v_0 \cos(\theta)\, t, \\ y(t) &= y_0 + v_0 \sin(\theta)\, t - \tfrac{1}{2} g t^2. \end{aligned}$$

The visualization numerically finds the time when $y(t)$ reaches the ground, and draws the path and impact point accordingly.

The default settings correspond exactly to this problem:

• $y_0 = 50$ m
• $v_0 = 20$ m/s
• $\theta = 60^\circ$
• $g \approx 9.8$ m/s²

You will see the ball arc upward and then fall, hitting the ground at about 5.4 seconds and 54 meters from the building, matching the analytic solution above.