Question

A ball is thrown from a cliff the cliff is 70 m high the ball reaches a maximum height of 100 m it reaches the ground at a .90 m from where it started

Answer 1

Let us write the equations of motion:

$x = v_0\cos\alpha t, \;\; y = y_0 + v_0\sin\alpha t - \dfrac{gt^2}{2}.$

So, $t = \dfrac{x}{v_0\cos\alpha}, \;\; y = y_0 + x\tan\alpha - \dfrac{g}{2v_0^2\cos^2\alpha}x^2$ .

The last equation is an equation of a parabola.

When x = 90, y = 0, when x = 0, y = 70, and the ordinate of a vertex is 100.

So, $y = 70 + x\tan\alpha - \dfrac{5}{v_0^2\cos^2\alpha}x^2$ .

$0 = 70 +90\tan\alpha - \dfrac{5}{v_0^2\cos^2\alpha}\cdot90^2$ (1)

The abscissa of a vertex is $x_0 = -\dfrac{\tan\alpha}{-2\frac{5}{v_0^2\cos^2\alpha}} = \dfrac{v_0^2\cos^2\alpha\tan\alpha}{10}.$ And $100 = 70 + \dfrac{v_0^2\cos^2\alpha\tan\alpha}{10}\tan\alpha - \dfrac{v_0^2\cos^2\alpha\tan^2\alpha}{20} = 70 + \dfrac{v_0^2\cos^2\alpha\tan^2\alpha}{20} .$ (2)

Therefore, (1) $0 = 7 +9\tan\alpha - \dfrac{5}{v_0^2\cos^2\alpha}\cdot810.$

(2) $30 = \dfrac{v_0^2\cos^2\alpha\tan^2\alpha}{20} \Rightarrow v_0^2 = \dfrac{600}{\cos^2\alpha\tan^2\alpha}$

(1) $0 = 7 +9\tan\alpha - \dfrac{5\cos^2\alpha\tan^2\alpha}{600\cos^2\alpha}\cdot810, \;\; 0 = 7+9\tan\alpha - \dfrac{27}{4}\tan^2\alpha.$

$\tan\alpha \approx 1.88, \alpha \approx 62^\circ.$

$v_0\approx$ 27.7 m/s.