Answer to Question #346278 in Physics for Jerry

Question

A stone is thrown at an initial speed of 125 ft/s directed 60 ° above the horizontal at the cliff of

height, y. The stone strikes the cliff 5.8 seconds after launching. Find the height of the cliff and

the speed of the stone just before impact at the top of the cliff.

Answer 1

$$y=v_0\sin\theta t-gt^2/2$$

$$h=125\sin60^\circ*5.8-32.17*5.8^2/2\\ =87\:\rm ft$$

$$v_x=v_0\cos\theta=125*\cos 60^\circ=62.5\:\rm ft/s$$

$$v_y=v_0\sin\theta-gt \\=125*\sin 60^\circ-32.17*5.8=-78.3\:\rm ft/s$$

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{62.5^2+(-78.3)^2}=100\:\rm ft/s$$