An object is projected from the top of a tower of height $H$ at an angle $\theta$ with the horizontal. It strikes the ground at $P$ lying at a distance $D$ from the foot of the tower. Calculate the maximum height attained by the object from the ground level:
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When standard formulas containing the initial velocity parameter $v$ cannot be directly evaluated because $v$ is not explicitly given, use the boundary conditions of the trajectory equation to eliminate $v$ and express the final result in terms of known geometric variables ($H, D, \theta$).
Step 1: Set up the picture with coordinates. Place the foot of the tower at the origin $(0,0)$. The object is launched from the top of the tower, so its starting point is $(0,H)$. It is thrown at angle $\theta$ with the horizontal and lands on the ground at a horizontal distance $D$.
Step 2: Write the path equation. A projectile follows a curved path described by: \[ y = H + x\tan\theta - \frac{g x^2}{2 v^2 \cos^2\theta} \] Here $y$ is the height and $x$ is the horizontal distance from the tower foot.
Step 3: Use the landing point to find the unknown. When the object lands, $y = 0$ and $x = D$. Putting these in: \[ 0 = H + D\tan\theta - \frac{g D^2}{2 v^2 \cos^2\theta} \] This lets us replace the awkward term $\dfrac{g}{2 v^2 \cos^2\theta}$ with known letters: \[ \frac{g}{2 v^2 \cos^2\theta} = \frac{H + D\tan\theta}{D^2} \]
Step 4: Find the position of greatest height. The object rises until its vertical speed becomes zero. In terms of the path, the top is reached where the height is largest. Using the trajectory, the extra rise above the launch point works out to: \[ h_{rise} = \frac{\tan^2\theta}{4} \cdot \frac{D^2}{H + D\tan\theta} \] This comes from finding the peak of the parabola.
Step 5: Add the tower height. The maximum height above the ground is the tower height $H$ plus this rise above the launch point: \[ H_{max} = H + \frac{D^2 \tan^2\theta}{4(H + D\tan\theta)} \]
Step 6: Match with the options. This exactly matches the second choice, so the maximum height from the ground is: \[ \boxed{H + \frac{D^{2}\tan^{2}\theta}{4(H + D\tan\theta)}} \]
