Answer to Question #345889 in Physics for seph

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

$$y=v_0\sin\theta t-gt^2/2$$ $$h=125\sin60^\circ*5.8-32.17*5.8^2/2\\ =87\:\rm ft$$ $$v_x=v_0\cos\theta=125*\cos 60^\circ=62.5\:\rm ft/s$$ $$v_y=v_0\sin\theta-gt \\=125*\sin 60^\circ-32.17*5.8=-78.3\:\rm ft/s$$ $$v=\sqrt{v_x^2+v_y^2}=\sqrt{62.5^2+(-78.3)^2}=100\:\rm ft/s$$