Question

A stone is projected from the ground level with an initial speed 𝑣0 directed 350 above the horizontal. Five seconds later, it lands on a cliff 8 m high.

a) What is the magnitude of the initial velocity of the stone?

b) What is the horizontal and vertical component of its velocity and velocity magnitude just before the stone hits the cliff?

c) How far from the release point horizontally does the stone hit the cliff?

Answer 1

Let y-axis be directed vertically and x-axis be directed horizontally towards the cliff.

$v_x = v_0\cos\alpha = v_0\cos35^\circ, \\ v_y = v_0\sin\alpha - gt = v_0\sin35^\circ -gt.$

The coordinates after time t will be

$x = v_0\cos\alpha\cdot t, \\ y = v_0\sin\alpha\cdot t - \dfrac{gt^2}{2}$ .

a) Therefore, $8 = v_0\sin35^\circ \cdot 5 - \dfrac{10\cdot 5^2}{2}, \;\; v_0 = 46.4\,\mathrm{m/s}.$

b) After 5 seconds the components of velocity and the magnitude of velocity will be

$v_x = v_0\cos 35^\circ = 38\,\mathrm{m/s}, \\ v_y = v_0\sin35^\circ - gt = -23.4\,\mathrm{m/s}; \\ v = \sqrt{v_x^2+v_y^2} = 44.6\,\mathrm{m/s}.$

c) The x-coordinate will be

$x = v_0\cos35^\circ\cdot 5 = 190\,\mathrm{m}.$