Question

A ball is thrown at 12m/s at an angle of 35 degrees above the horizontal. a) Find its velocity 1.0s later.b) At what time after it was thrown will the ball be headed at an angle of 20 degrees above the horizontal? c) At 20 degrees below the horizontal

Answer 1

Question 15509

Let $\varphi_0$ denote an angle at which the ball is thrown, and $\nu_0$ denote the initial velocity, $\varphi$ denote the angle between the horizontal line and velocity vector.

a) The general equation for velocity of an object, moving with constant acceleration is

$$\vec{v} = \vec{v}_0 + \vec{a}t$$

. Lets write it in a vector form: $\begin{pmatrix} v_x \\ v_y \end{pmatrix} = \begin{pmatrix} v_{0x} \\ v_{0y} \end{pmatrix} + \begin{pmatrix} 0 \\ -g \end{pmatrix} t = \begin{pmatrix} v_0 \cos \varphi \\ v_0 \sin \varphi \end{pmatrix} + \begin{pmatrix} 0 \\ -g \end{pmatrix} t$ , where

$$g = 9.81 \, \text{m}/\text{s}^2$$

. From here, for $t = 1$ , $\nu_x \approx 9.83 \, \text{m}/\text{s}, \nu_y \approx -2.91 \, \text{m}/\text{s}$ .

b) The angle might be found from angle between $\nu_x$ and $\nu_y$ : $\tan \varphi = \frac{\nu_x(t)}{\nu_x(t)} = \frac{\nu_0 \sin \varphi_0 - g t}{\nu_0 \cos \varphi_0}$ ,

which gives $t_1 = \frac{v_0}{g} (\sin \varphi_0 - \cos \varphi_0 \tan \varphi) \approx 0.34 \, \text{s}$ .

c) Below the horizontal means formal substitution $\varphi \rightarrow -\varphi$ , which gives

$$t_2 = \frac{v_0}{g} (\sin \varphi_0 + \cos \varphi_0 \tan \varphi) \approx 1.07 \, \text{s}.$$