Projectile from a Building: Flight Path and Impact Point

We model the ball as a projectile launched from a height with an initial speed and angle, under constant gravitational acceleration.

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1. Problem setup

• 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 decompose the initial velocity:

$$\begin{aligned} v_{0x} &= v_0 \cos\theta \\ v_{0y} &= v_0 \sin\theta \end{aligned}$$

With $v_0 = 20$ and $\theta = 60^\circ$:

$$\cos 60^\circ = \tfrac12,\quad \sin 60^\circ = \tfrac{\sqrt{3}}{2}$$

So:

$$\begin{aligned} v_{0x} &= 20 \cdot \tfrac12 = 10\,\text{m/s} \\ v_{0y} &= 20 \cdot \tfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.32\,\text{m/s} \end{aligned}$$

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2. Position as a function of time

We place the origin at ground level directly below the launch point:

• $x(t)$ horizontal position
• $y(t)$ vertical position (0 at the ground, positive upward)

Equations of motion:

$$\begin{aligned} x(t) &= v_{0x} t = 10 t \\ y(t) &= h_0 + v_{0y} t - \tfrac12 g t^2 \\ &= 50 + 10\sqrt{3}\, t - 4.9 t^2 \end{aligned}$$

We want the impact time when the ball hits the ground, i.e. $y(t) = 0$.

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3. Solving for impact time

Set

$$50 + 10\sqrt{3}\, t - 4.9 t^2 = 0$$

Rewriting in standard quadratic form $at^2 + bt + c = 0$:

$$-4.9 t^2 + 10\sqrt{3}\, t + 50 = 0$$

Multiply by $-1$ for convenience:

$$4.9 t^2 - 10\sqrt{3}\, t - 50 = 0$$

Here:

$$\begin{aligned} a &= 4.9 \\ b &= -10\sqrt{3} \\ c &= -50 \end{aligned}$$

Using the quadratic formula:

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

Compute discriminant:

$$\begin{aligned} b^2 &= (-10\sqrt{3})^2 = 100 \cdot 3 = 300 \\ 4ac &= 4 \cdot 4.9 \cdot (-50) = -980 \\ b^2 - 4ac &= 300 - (-980) = 1280 \end{aligned}$$

So:

$$t = \frac{10\sqrt{3} \pm \sqrt{1280}}{2 \cdot 4.9}$$

We only accept the positive root (time must be positive). Numerically:

• $\sqrt{1280} \approx 35.78$
• $10\sqrt{3} \approx 17.32$
• Numerator (positive root): $17.32 + 35.78 \approx 53.10$
• Denominator: $2\cdot 4.9 = 9.8$

Thus:

$$t_\text{hit} \approx \frac{53.10}{9.8} \approx 5.42\,\text{s}$$

So the ball hits the ground approximately 5.4 seconds after launch.

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4. Horizontal distance at impact

Horizontal motion is uniform (no horizontal acceleration):

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

At $t = t_\text{hit} \approx 5.42$:

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

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

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5. Summary of results

• Time of impact: $t_\text{hit} \approx 5.4\,\text{s}$ after launch.
• Horizontal distance from building: $x_\text{hit} \approx 54\,\text{m}$.

The visualization below lets you vary the initial speed, launch angle, and building height, and watch how the trajectory and impact point change in real time. The neon path shows the projectile's arc; a small marker shows the current position at time $t$, and a glowing point marks the impact location when it hits the ground ($y = 0$).