Answer to Question #93838 in Mechanics | Relativity for ashley

Question

A ball rolls off a 108 cm high table at 5.6 m/s. What is its speed when it hits the ground, in m/s?

A baseball is hit at an angle of 26.3° and velocity of 26.2 m/s on level ground. A stiff head wind causes a horizontal acceleration of 1.97 m/s2 opposing the ball's motion. How far in meters does the ball travel on level ground?

You are the quarterback ready to pass the ball to a receiver already 23.5 m down the field (from you). He is running straight away at a constant 6.31 m/s, and you will release the ball at an angle of 45.9°. With what speed (in m/s) should you throw the ball?

Answer 1

1.

$$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.08)}{9.8}}=0.469\ s$$

$$v=\sqrt{(v_x)^2+(v_y)^2}=\sqrt{(v_x)^2+(gt)^2}$$

$$v=\sqrt{(5.6)^2+((9.8)(0.469))^2}=7.2\frac{m}{s}$$

2.

$$t=\frac{2v\sin{26.3°}}{g}=\frac{2(26.2)\sin{26.3°}}{9.8}=2.369\ s$$

$$s=v\cos{26.3°}t-0.5at^2$$

$$s=26.2\cos{26.3°}(2.369)-0.5(1.97)(2.369)^2=50.1\ m$$

3.

$$x=vt+s$$

$$t=\frac{2V\sin{45.9°}}{g}$$

$$x=V \cos{45.9°}t$$

$$V \cos{45.9°}t=vt+s$$

$$V \cos{45.9°}\frac{2V\sin{45.9°}}{g}=v\frac{2V\sin{45.9°}}{g}+s$$

$$V \cos{45.9°}\frac{2V\sin{45.9°}}{9.8}=6.31\frac{2V\sin{45.9°}}{9.8}+23.5$$

$$V=20.4\frac{m}{s}$$