An object thrown at some angle to the horizontal lands back…

An object thrown at some angle to the horizontal lands back on level ground after $T = 6\ \mathrm{s}$. How high did it rise?

Explanation of Method

Maximum height from the flight time.

Step 1: Vertical launch speed

\[ T = \frac{2v_0\sin\theta}{g} \;\Longrightarrow\; v_0\sin\theta = \frac{gT}{2} = 29.43\ \mathrm{m/s}. \]

Step 2: Maximum height

\[ H = \frac{(v_0\sin\theta)^2}{2g} = \frac{g T^2}{8} = \frac{9.81 \cdot 36}{8}. \]

\[H \approx 44.1\ \mathrm{m}\]

$y\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $3$ $6$ $44.1$ $H = \tfrac{g T^2}{8} \approx 44.1\ \mathrm{m}$ $\text{up: } 3\ \mathrm{s}$ $\text{down: } 3\ \mathrm{s}$

Height against time. The top is reached halfway, after 3 s; falling from there for 3 s covers $\tfrac12 g \cdot 3^2 \approx 44\ \mathrm{m}$, whatever the launch angle.

Result

\[\boxed{\,H \approx 44.1\ \mathrm{m}\,}\]

Remark

Note. The speed and the angle are not needed separately: the height depends only on the flight time. The top is reached after $T/2 = 3\ \mathrm{s}$, and falling freely for 3 s covers $\tfrac12 g \cdot 9 = 44.1\ \mathrm{m}$.