Q55P - University Physics with Modern Physics

Question

A baseball thrown at an angle of  $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$ above the horizontal strikes a building 18.0 m away at a point  8.0 m above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball’s initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

Step-by-Step Solution

Answer

1. The initial velocity magnitude is 16.55 m/s . 
2. The direction is below the horizontal at  $\theta =40.64°$ and  $10.90 m/s$ is the magnitude of velocity when the ball just strike.

Step 1: A concept of the equation of motion:

The velocity contains two components, one is a horizontal component and another one is vertical. 

 

According to Newton’s laws of motion,

 

$\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$ 

 

Here,  $s,u,t,$ and  a are displacement, initial velocity, time, and acceleration respectively.

Step 2: Consider the given data:

Angle,  $\theta =60°$

 

Horizontal distance,  $u=18 m$

 

Vertical distance,  $v=8 m$

Step 3: (a) Initial velocity’s magnitude:

For horizontal motion, gravity is zero.

                                                                               

 $$\begin{gathered} s_H=u\cos \theta t \\ t=\frac{s_H}{u \cos \theta } \end{gathered}$$

 

For the vertical motion, gravity is negative

 

 $$\begin{gathered} s_v=u \sin \theta t-\frac{1}{2}g{t}^2 \\ =u \sin \theta \left(\frac{s_H}{u \cos \theta }\right)-\frac{1}{2}g{\left(\frac{s_H}{u \cos \theta }\right)}^2 \\ 8 m =18 m\times \tan 60°-\frac{1}{2}9.8 m/s^2{\left(\frac{18 m}{u \cos {60}^{°}}\right)}^2 \\ 8 m =18 m\times 1.73-\left(\frac{6350.4}{u^2}\right) \\ \left(\frac{6350.4}{u^2}\right)=23.14 \\ u=16.56 m/s \\ {s}_v=u \sin \theta \left(\frac{s_H}{u \cos \theta }\right)-\frac{1}{2}g{\left(\frac{s_H}{u \cos \theta }\right)}^2 \\ 8 m=18 m\times \tan 60°-\frac{1}{2}\left(9.8 m/s^2\right){\left(\frac{18 m}{u \cos 60°}\right)}^2 \\ 8m =\left(31.176 m\right)-\left(\frac{6,350.4}{u^2}\right) \\ \frac{6,350.4}{u^2}=23.176 \\ {u}^2=274.0075 m^2/s^2 \\ u=16.55 m/s \end{gathered}$$                                                                                                This is the initial velocity magnitude.

Step 4: (b) Magnitude and direction of the velocity of the ball just before strike

The final horizontal velocity component,

 

$$\begin{gathered} u \cos \theta =16.55 m/s\times \cos 60° \\ =8.27 m/s \end{gathered}$$      

 

The final vertical velocity is v, from the Newton’s laws of motion,

 

 $$\begin{gathered} v^2=u^2+2gs \\ v=\sqrt{{\left(u\sin \theta \right)}^2+2gs} \\ =\sqrt{\left(14.4 m/s^2\right)+2\times -9.8 m/s^2\times 8 m} \\ =7.1 m/s \end{gathered}$$

 

So the magnitude of velocity when the ball just strike,

                                                                                                                                                             

$$\begin{gathered} v_{final}=\sqrt{{\left(8.27 m/s\right)}^2+{\left(7.1 m/s\right)}^2} \\ =10.90 m/s \end{gathered}$$     

 

For direction,

 

$$\begin{gathered} \tan \theta =\frac{7.1}{8.27} \\ \theta =40.64° \end{gathered}$$       

 

The direction is below the horizontal at $\theta =40.64°$ .