(a) A girl threw a stone horizontally with a velocity of 30m/s from the top of a cliff 50m high. How far from the foot of the cliff does the stone strike th...

Answer Details

(a) Horizontal projectile from a cliff

Horizontally the stone travels at a constant \(30\,\text{m/s}\); vertically it starts with zero vertical velocity and falls under gravity. First find the time of flight from the vertical motion, using \(h=\tfrac{1}{2}gt^{2}\):

\[50=\tfrac{1}{2}(10)t^{2}\;\Rightarrow\; 50=5t^{2}\;\Rightarrow\; t^{2}=10\;\Rightarrow\; t=\sqrt{10}\approx 3.16\,\text{s}.\]

The horizontal distance (range) is

\[R=\text{(horizontal speed)}\times t=30\sqrt{10}\approx 94.9\,\text{m}.\]

The stone lands about \(94.9\,\text{m}\) from the foot of the cliff.

(b) Body in equilibrium on two strings

The weight of the body is \(W=mg=2\times 10=20\,\text{N}\), acting vertically downward. String \(AP\) (tension \(T_1\)) makes \(56^{\circ}\) with the upward vertical, and string \(AR\) (tension \(T_2\)) is horizontal. Resolve the forces at \(A\).

Vertical equilibrium: only \(T_1\) has a vertical component, and it supports the weight:

\[T_1\cos 56^{\circ}=20\;\Rightarrow\; T_1=\frac{20}{\cos 56^{\circ}}=\frac{20}{0.5592}\approx 35.8\,\text{N}.\]

Horizontal equilibrium: the horizontal component of \(T_1\) is balanced by \(T_2\):

\[T_2=T_1\sin 56^{\circ}=20\tan 56^{\circ}=20(1.4826)\approx 29.7\,\text{N}.\]

Hence \(T_1\approx 35.8\,\text{N}\) and \(T_2\approx 29.7\,\text{N}\).