Two stones are projected from the top of a cliff h meters high, with the same speed u so as to hit the ground

Question

Two stones are projected from the top of a cliff h meters high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle θ to the horizontal then θ equals:

Accepted Answer

For the horizontally projected stone:
Time

t_{1} = \frac{2 h}{g}

.
Horizontal distance

x = u t_{1} = u \frac{2 h}{g}

.
For the stone projected at angle

θ

:
Horizontal distance

x = (u cos θ) t_{2} ⟹ t_{2} = \frac{x}{u c o s θ} = \frac{2 h / g}{c o s θ}

.
Vertical displacement

y = - h = (u sin θ) t_{2} - \frac{1}{2} g t_{2}^{2}

.

- h = u sin θ (\frac{x}{u cos θ}) - \frac{1}{2} g (\frac{x ^{2}}{u ^{2} cos ^{2} θ})

- h = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}

Substitute

x^{2} = \frac{2 h u ^{2}}{g}

:

- h = x tan θ - \frac{g}{2 u ^{2}} (\frac{2 h u ^{2}}{g}) sec^{2} θ = x tan θ - h (1 + tan^{2} θ)

- h = x tan θ - h - h tan^{2} θ ⟹ h tan^{2} θ = x tan θ

Since

θ \neq = 0

:

tan θ = \frac{x}{h} = \frac{u 2 h / g}{h} = u \frac{2}{g h}

.

Two stones are projected from the top of a cliff $h$ meters high, with the same speed $u$ so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle $θ$ to the horizontal then $tan θ$ equals:

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Two stones are projected from the top of a cliff h meters high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle θ to the horizontal then tanθ equals: