Answer to Question #94771 in Mechanics | Relativity for Ikegreat

Question

A stone is thrown from the top of a building upwards at an angle of 30.0 to the horizontal with an initial speed of 20.0m/s. The height of the building is 45m.
1. How long does it take to reach the ground?.
2. What is the speed of the stone just before it strikes the ground?

Answer 1

The equations of the motion of a stone

$$x(t)=v_0\cos \theta\cdot t\\ y(t)=H+v_0\sin\theta\cdot t-\frac{gt^2}{2}$$

$$x(t)=20\cos 30^{\circ}\cdot t\\ y(t)=45+20\sin30^{\circ}\cdot t-5t^2$$

(a) When a stone strikes the ground $y=0,$ so

$$45+20\sin30^{\circ}\cdot t-5t^2=0$$

Root

$$t=4.16\:\rm s$$

(b) The final speed of a stone

$$v=\sqrt{(v_0\cos \theta)^2+(v_0\sin \theta-gt)^2}$$

$$v=\sqrt{(20\cos30^{\circ})^2+(20\sin 30^{\circ}-10\times 4.16)^2}$$

$$=36\:\rm m/s$$