Projectile From a Building: When and Where Does It Land?

We model the ball as a projectile launched from the top of a 50 m building with initial speed 20 m/s at an angle of 60° above the horizontal.

We assume constant gravitational acceleration:

$$g \approx 9.8\, \text{m/s}^2$$

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1. Decompose the initial velocity

Initial speed: $v_0 = 20\,\text{m/s}$

Launch angle: $\theta = 60^\circ$

Horizontal and vertical components:

$$\begin{aligned} v_{0x} &= v_0 \cos\theta = 20 \cos 60^\circ = 20 \cdot 0.5 = 10\,\text{m/s},\\[4pt] v_{0y} &= v_0 \sin\theta = 20 \sin 60^\circ = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32\,\text{m/s}. \end{aligned}$$

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2. Position as a function of time

Let $x(t)$ be horizontal position (meters) and $y(t)$ be height above ground (meters). Take:

• $x = 0$ directly under the launch point,
• $y = 0$ at ground level,
• $y(0) = h = 50\,\text{m}$.

Equations of motion:

$$\begin{aligned} x(t) &= v_{0x} t = 10 t,\\[4pt] y(t) &= h + v_{0y} t - \tfrac{1}{2} g t^2 = 50 + 17.32\,t - 4.9 t^2. \end{aligned}$$

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3. Find the impact 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 in $t$:

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

Multiply by $-1$:

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

Use the quadratic formula:

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

Compute the discriminant:

$$\begin{aligned} D &= 17.32^2 + 4\cdot 4.9 \cdot 50 \\ &\approx 299.98 + 980 \\ &\approx 1279.98. \end{aligned}$$

$\sqrt{D} \approx 35.78$. So:

$$\begin{aligned} t &= \frac{17.32 \pm 35.78}{9.8}. \end{aligned}$$

The two solutions:

• $t_1 = \dfrac{17.32 - 35.78}{9.8} \lt 0$ (negative time, not physical for our situation),
• $t_2 = \dfrac{17.32 + 35.78}{9.8} \approx \dfrac{53.10}{9.8} \approx 5.42\,\text{s}.$

So the impact time is:

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

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4. Find the horizontal range (where it lands)

Horizontal position at impact:

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

So the ball hits the ground about 54 meters horizontally away from the point directly below the launch point.

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

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5. Summary

• Time until the ball hits the ground: $\approx 5.4\,\text{s}$.
• Horizontal distance from the base of the building: $\approx 54\,\text{m}$.

This visualization lets you adjust the building height, launch speed, angle, and gravity, and see how the trajectory, impact time, and range change in real time.