Question

Shots fired simultaneously from the top and bottom of a vertical cliff with the elevation $\alpha$ and $\beta$ respectively, strike an object simultaneously at the same point on the ground. If s is the horizontal distance of the object from the cliff, then what is the height of the cliff. ($\beta > \alpha$)

Answer 1

$h = s(\tan\beta - \tan\alpha)$

Answer 2

Let the height of the cliff be $h$ and the horizontal distance from the cliff to the object be $s$. Two projectiles are fired simultaneously:

• From the top of the cliff with initial speed $u$ at an angle $\alpha$ with the horizontal.
• From the bottom of the cliff with initial speed $v$ at an angle $\beta$ with the horizontal where $\beta > \alpha$.

For the projectile from the top:

1. Horizontal motion:

   $$s = u \cos\alpha \, t \quad \Rightarrow \quad u = \frac{s}{t\cos\alpha}$$

2. Vertical motion (starting from height $h$ and landing at ground level):

   $$h + u \sin\alpha\, t - \frac{1}{2}gt^2 = 0 \quad \Rightarrow \quad h = \frac{1}{2}gt^2 - s \tan\alpha$$

For the projectile from the bottom:

1. Vertical motion (starts from ground and returns to ground):

   $$v \sin\beta\, t - \frac{1}{2}gt^2 = 0 \quad \Rightarrow \quad t = \frac{2v\sin\beta}{g}\quad (\text{ignoring } t=0)$$

2. Horizontal motion:

   $$s = v \cos\beta\, t = v \cos\beta \cdot \left(\frac{2v\sin\beta}{g}\right) = \frac{v^2\sin2\beta}{g}$$

   This gives:

   $$v^2 = \frac{g\,s}{\sin2\beta}$$

Since both projectiles hit the target simultaneously, the same time $t$ holds for both. Substitute the expression for $t$ obtained from the bottom shot into the vertical equation of the top shot.

Calculate:

$$t = \frac{2v\sin\beta}{g} \quad \Rightarrow \quad t^2 = \frac{4v^2\sin^2\beta}{g^2}$$

and using $v^2 = \frac{g s}{\sin2\beta}$:

$$t^2 = \frac{4\sin^2\beta}{g^2}\cdot\frac{g s}{\sin2\beta} = \frac{4s\sin^2\beta}{g\,\sin2\beta}$$

Now, substitute into the top projectile’s vertical equation:

$$h = \frac{1}{2}g t^2 - s \tan\alpha = \frac{1}{2}g\left(\frac{4s\sin^2\beta}{g\,\sin2\beta}\right) - s \tan\alpha$$ $$h = \frac{2s\sin^2\beta}{\sin2\beta} - s \tan\alpha$$

Recall the trigonometric identity:

$$\sin2\beta = 2\sin\beta\cos\beta \quad \Rightarrow \quad \frac{\sin^2\beta}{\sin2\beta} = \frac{\sin^2\beta}{2\sin\beta\cos\beta} = \frac{\sin\beta}{2\cos\beta} = \frac{1}{2}\tan\beta$$

Thus,

$$h = 2s \cdot \left(\frac{1}{2}\tan\beta\right) - s \tan\alpha = s\tan\beta - s\tan\alpha$$ $$h = s\left(\tan\beta - \tan\alpha\right)$$