A projectile leaves with speed v_0 = 400 m/s at an angle

A projectile leaves with speed $v_{0} = 400\ \text{m/s}$ at an angle $\alpha$ to the horizon. Find the height at which it strikes a vertical wall $3000\ \text{m}$ away, and its speed at that moment. Air resistance is neglected.

Step 1: The equations of motion

From Problem 344, with $g = 9.81\ \text{m/s}^{2}$: $x = v_{0}t\cos\alpha, \qquad y = v_{0}t\sin\alpha-\frac{gt^{2}}{2}$.

Step 2: Time of impact

The wall is reached when $x = 3000$: \[3000 = 400\,t\cos\alpha \;\Longrightarrow\; t^{*} = \frac{3000}{400\cos\alpha} = \frac{7.5}{\cos\alpha}\ \text{s}.\]

Step 3: Height at impact

The first term gives $400\sin\alpha\cdot\dfrac{7.5}{\cos\alpha} = 3000\tan\alpha$. The second is

\[ \frac{g}{2}\left(\frac{7.5}{\cos\alpha}\right)^{2} = \frac{9.81}{2}\cdot\frac{56.25}{\cos^{2}\alpha} = \frac{275.906}{\cos^{2}\alpha}. \]

Using $\dfrac{1}{\cos^{2}\alpha} = 1+\tan^{2}\alpha$: $y^{*} = 3000\tan\alpha-275.906\bigl(1+\tan^{2}\alpha\bigr)\ \text{m}$.

Step 4: Velocity components at impact

The horizontal component never changes; the vertical one has lost $gt^{*}$. Note that $gt^{*} = 9.81\cdot\dfrac{7.5}{\cos\alpha}$, so the numerator is $9.81\times7.5 = 73.575$ — not $9.81\times15$:

\[ v_{x} = 400\cos\alpha, \qquad v_{y} = 400\sin\alpha-\frac{73.575}{\cos\alpha}. \]

Step 5: Speed at impact

Squaring and adding, the two $400^{2}$ terms combine by the Pythagorean identity: $v^{2} = 160\,000-2\cdot400\cdot73.575\tan\alpha+\frac{73.575^{2}}{\cos^{2}\alpha}$.

\[v = \sqrt{160\,000-58\,860\tan\alpha+5413.28\bigl(1+\tan^{2}\alpha\bigr)}\ \frac{\text{m}}{\text{s}}\]

Step 6: Verification at $\alpha = 45^{\circ}$

\[t^{*} = \frac{7.5}{0.70711} = 10.6066\ \text{s}; \qquad y^{*} = 3000-275.906\cdot2 = 2448.19\ \text{m}\]

\[\text{Direct: } 400\cdot10.6066\cdot0.70711-4.905\cdot10.6066^{2} = 3000-551.81 = 2448.19 \;\;\checkmark\]

\[v_{x} = 282.84,\ v_{y} = 282.84-104.06 = 178.78,\ v = \sqrt{282.84^{2}+178.78^{2}} = 334.6 \;\;\checkmark\]

The closed formula agrees: $\sqrt{160\,000-58\,860+10\,826.6} = \sqrt{111\,967} = 334.6\ \text{m/s}$.

Step 7: Reading the answer

The height $y^{*}$ is a downward parabola in $\tan\alpha$, greatest when $\tan\alpha = \dfrac{3000}{2\cdot275.906} \approx 5.437$, i.e. $\alpha \approx 79.6^{\circ}$; it is positive only between the two roots, which are the angles that actually clear the wall.