Question

A rock is thrown horizontally from the top of a cliff 88 m high, with a horizontal speed of 25 m/s. with what velocity does the rock hit

Answer 1

$$x(t)=v_{0x}t$$

$$y(t)=h+0\cdot t-\dfrac{gt^2}{2}$$

$$v_y(t)=y'(t)=-gt$$

The rock hits the land: $y(t_1)=0.$

$$h-\dfrac{gt_1^2}{2}=0, t_1\ge0$$

$$t_1=\sqrt{\dfrac{2h}{g}}$$

$$v_y(t_1)=-g\sqrt{\dfrac{2h}{g}}=-\sqrt{2gh}$$

$$|v(t_1)|=\sqrt{v_{0x}^2+(-\sqrt{2gh})^2}$$

$$=\sqrt{v_{0x}^2+2gh}$$

$$=\sqrt{(25m/s)^2+2(9.81m/s^2)(88m)}$$

$$=48.493m/s$$

$$\tan \theta=-\dfrac{\sqrt{2gh}}{v_{0x}}=-\dfrac{\sqrt{2(9.81m/s^2)(88m)}}{25m/s}$$

$$\approx-1.662$$

$$\theta\approx121\degree$$

The rock hits the land with the velocity 48.493 m/s at the angle $121\degree$ with respect to the ground.