Question

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:

• A. $\frac{v^{2}\sin^{2}\theta}{2g}$
• B. $H+\frac{D^{2}\tan^{2}\theta}{4(H+D \tan\theta)}$
• C. $H+D\tan\theta$
• D. $H+\frac{D^{2}}{v \cos\theta}$

Hint:

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 ($H, D, \theta$).

Answer 1

The Correct Option is B

Concept: The motion of a projectile launched from an elevated point can be analyzed by separating it into two orthogonal spatial coordinates. Let the base of the tower be the origin $(0,0)$. The coordinates of the launch point at the top of the tower are $(0, H)$. The initial velocity vector is given by: $$u_x = v \cos\theta, \quad u_y = v \sin\theta$$ The trajectory equation for a projectile tracking path positions $(x, y)$ is given by: $$y = H + x\tan\theta - \frac{gx^2}{2v^2\cos^2\theta}$$

Step 1: The problem specifies that the object strikes the ground at point $P$, which is situated at a distance $D$ from the foot of the tower. Thus, the coordinate point $(D, 0)$ must satisfy the equation of the trajectory: $$0 = H + D\tan\theta - \frac{gD^2}{2v^2\cos^2\theta}$$ Rearranging this terms to isolate the factor containing the velocity and gravity configurations: $$\frac{gD^2}{2v^2\cos^2\theta} = H + D\tan\theta$$ From this equation, we can express the term $\frac{g}{2v^2\cos^2\theta}$ directly as: $$\frac{g}{2v^2\cos^2\theta} = \frac{H + D\tan\theta}{D^2} \quad \cdots (1)$$

Step 2: The standard formula for the maximum height attained by a projectile *above its point of projection* ($h_{\max}$) depends purely on its vertical component of initial velocity ($u_y = v\sin\theta$): $$h_{\max} = \frac{u_y^2}{2g} = \frac{v^2\sin^2\theta}{2g}$$ We can rewrite $h_{\max}$ by systematically grouping the terms to incorporate the expression derived in equation (1): $$h_{\max} = \frac{v^2\sin^2\theta}{2g} \cdot \frac{\cos^2\theta}{\cos^2\theta} = \frac{\sin^2\theta}{\cos^2\theta} \cdot \frac{v^2\cos^2\theta}{2g} = \tan^2\theta \cdot \left( \frac{v^2\cos^2\theta}{2g} \right)$$ Notice that the term in the parenthesis is the reciprocal of $\frac{2g}{v^2\cos^2\theta}$. Let us write: $$h_{\max} = \tan^2\theta \cdot \frac{1}{4 \cdot \left(\frac{g}{2v^2\cos^2\theta}\right)}$$

Step 3: Now substitute the expression for $\frac{g}{2v^2\cos^2\theta}$ from equation (1) into this rearranged form: $$h_{\max} = \tan^2\theta \cdot \frac{1}{4 \cdot \left( \frac{H + D\tan\theta}{D^2} \right)}$$ $$h_{\max} = \frac{D^2\tan^2\theta}{4(H + D\tan\theta)}$$

Step 4: The total maximum height reached measured directly from the ground baseline level ($H_{\text{total}}$) is the height of the tower plus the additional peak displacement scaled above the launch pad: $$H_{\text{total}} = H + h_{\max}$$ $$H_{\text{total}} = H + \frac{D^2\tan^2\theta}{4(H + D\tan\theta)}$$ This precisely matches the mathematical layout presented in Option (B).