Projectile from a Building – Impact Time and Range

We model the ball as a projectile launched from a 50 m high building with an initial speed of 20 m/s at a $60^\circ$ angle above the horizontal. Gravity acts downward with acceleration $g \approx 9.8\,\text{m/s}^2$.

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1. Setting up the motion equations

Let the launch point be at horizontal position $x = 0$ and vertical position $y = 50$ meters. Decompose the initial velocity:

• $v_0 = 20\,\text{m/s}$
• Angle $\theta = 60^\circ$

Horizontal and vertical components:

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

Using $\cos 60^\circ = \tfrac12$ and $\sin 60^\circ = \tfrac{\sqrt{3}}{2} \approx 0.866$:

$$\begin{aligned} v_{0x} &= 20 \cdot \tfrac12 = 10\;\text{m/s}\\ v_{0y} &= 20 \cdot 0.866 \approx 17.32\;\text{m/s} \end{aligned}$$

The position as a function of time $t$ (in seconds):

• Horizontal: $x(t) = v_{0x}\, t = 10t$
• Vertical: $y(t) = y_0 + v_{0y} t - \tfrac12 g t^2 = 50 + 17.32 t - 4.9 t^2$

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2. Time when the ball hits the ground

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

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

Rewriting as a standard quadratic:

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

Multiply by $-1$:

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

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

$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\quad a = 4.9,\; b = -17.32,\; c = -50.$$

So

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

Compute inside the square root:

$$\begin{aligned} (-17.32)^2 &\approx 299.7,\\ 4ac &= 4\cdot 4.9 \cdot (-50) = -980,\\ \Rightarrow b^2 - 4ac &\approx 299.7 - (-980) = 1279.7. \end{aligned}$$

$\sqrt{1279.7} \approx 35.78.$

Then

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

This gives two solutions:

1. $t_1 = \dfrac{17.32 - 35.78}{9.8} < 0$ (negative time, not physical),
2. $t_2 = \dfrac{17.32 + 35.78}{9.8} = \dfrac{53.10}{9.8} \approx 5.42\,\text{s}.$

We keep the positive time:

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

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3. Horizontal distance when it hits the ground

At impact, the horizontal position is

$$x_\text{impact} = v_{0x} t_\text{impact} = 10 \times 5.42 \approx 54.2\,\text{m}.$$

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

$$\boxed{x_\text{impact} \approx 54\,\text{m from the building base}}.$$

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4. What the visualization shows

This visualization treats the building top as the launch point and draws the full parabolic trajectory until it reaches the ground ($y = 0$). You can:

• Adjust the initial speed, angle, and building height to see how the time of flight and impact point change.
• See the moving ball following the computed path.
• Observe the highlighted impact point where the trajectory meets the ground.

A time slider lets you scrub through the motion, or you can enable auto-play to watch the motion in real time. The curve and markers update based on the exact projectile equations above:

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

The impact time is found numerically (or by the quadratic formula) as the smallest positive $t$ for which $y(t) = 0$.