A body is thrown horizontally from a height h = 6 m above…

A body is thrown horizontally from a height $h = 6\ \mathrm{m}$ above level ground and lands $l = 10\ \mathrm{m}$ away horizontally. Find (a) its initial speed; (b) the angle at which it hits the ground; (c) the equation of its trajectory.

Explanation of Method

Initial speed, impact angle and path of a horizontal throw.

Step 1: (a) Initial speed

\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{12}{9.81}} \approx 1.106\ \mathrm{s}, \]

\[ v_0 = \frac{l}{t} = l\sqrt{\frac{g}{2h}} = \frac{10}{1.106}. \]

\[v_0 \approx 9.04\ \mathrm{m/s}\]

Step 2: (b) Impact angle

$v_y = gt$ and $v_x = l/t$, so $\tan\theta = \dfrac{gt^2}{l} = \dfrac{2h}{l}$: $\tan\theta = \frac{2 \cdot 6}{10} = 1.2$.

\[\theta \approx 50.2^\circ \ \text{below the horizontal}\]

Step 3: (c) Trajectory

With $x$ horizontal and $y$ measured downward from the release point: $x = v_0 t$, $y = \tfrac12 g t^2$, so

\[ y = \frac{g}{2v_0^2}\,x^2 = \frac{h}{l^2}\,x^2. \]

\[y = 0.06\,x^2\ \ (x, y \text{ in metres, } y \text{ downward})\]

$h = 6\ \mathrm{m}$ $l = 10\ \mathrm{m}$ $50.2^\circ$ $v_0 \approx 9.04\ \mathrm{m/s}$ $y = 0.06\,x^2\ \ (y \text{ downward})$

Drawn to scale. The tangent at the landing point (dashed) meets the ground at 50.2°; its slope $2h/l = 1.2$ is twice the average slope of the drop.

Result

\[\boxed{\,v_0 \approx 9.04\ \mathrm{m/s}\,}\]

\[\boxed{\,\theta \approx 50.2^\circ \ \text{below the horizontal}\,}\]

\[\boxed{\,y = 0.06\,x^2\ \ (x, y \text{ in metres, } y \text{ downward})\,}\]

Remark

Note. The exact coefficient is $h/l^2 = 6/100 = 0.06$. The original values 9.01 m/s, 50.4° and 0.0605 differ slightly because the time was rounded to 1.11 s. Check: at $x = 10$, $y = 0.06 \cdot 100 = 6\ \mathrm{m}\ \checkmark$