Q17E - University Physics with Modern Physics

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30 m/s and at an angle of $\mathbf{36}\mathbf{.}\mathbf{9}\mathbf{°}$ above the horizontal. Ignore air resistance. (a) At what two times is the baseball at a height of 10 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball’s velocity at each of the two times calculated in part (a). (c) What are the magnitude and direction of the baseball’s velocity when it returns to the level at which it left the bat?

Step-by-Step Solution

Answer

a) The two times for ball to elapse is 2.991 s and 0.683 s.

b) The horizontal component of velocity is not change and vertical component of the velocity at $t_1$ is 11.3 m/s , and vertical component of velocity at $t_2$ is $-11.3 \mathrm{m}/\mathrm{s}$.

c) The velocity of the ball when it returns is 30 m/s and the direction of the velocity of the ball is downwards in horizontal direction at the angle of $-36.9°$.

Step 1: Identification of given data

The given data can be listed below,

• The speed of the bat is, $v_0=30 \mathrm{m}/\mathrm{s}$.
• The angle at which ball hit is, $\theta =36.9°$.
• The height gain by baseball is, $h=10 \mathrm{m}$.

Step 2: Concept/Significance of projectile.

Any item that is given an initial velocity and subsequently pursues a route wholly defined by gravitational acceleration is referred to as a projectile.

Step 3: (a) Determination of two times is the baseball at a height of above the point at which it left the bat.

The displacement of the ball is given by,

$y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$

Here, $v_{0y}$ is the vertical component of velocity whose value is $v_0\sin \theta$, $t$ is the elapsed time, $a_y$ is the acceleration in vertical direction whose value is $-g$.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{y}=\left({\mathrm{v}}_0\mathrm{sin\theta }\right)\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ 10 \mathrm{m}=\left(30 \mathrm{m}/\mathrm{s}\right)\left(\sin 36.9°\right)\mathrm{t}-\frac{1}{2}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right){\mathrm{t}}^2 \\ 0=\left(4.90 \mathrm{m}/\mathrm{s}\right){\mathrm{t}}^2-\left(18 \mathrm{m}/\mathrm{s}\right)\mathrm{t}+10 \mathrm{m} \\ \mathrm{t}=\frac{1}{9.8}+[18\pm \sqrt{{\left(-18\right)}^2-4\left(4.90\right)\left(10\right)}]\mathrm{s} \\ =\left(1.837\pm 1.154\right)\mathrm{s} \end{gathered}$$ 

From above solution there are two solutions for elapsed time which is given by,

$$\begin{gathered} t_1=\left(1.837+1.154\right)s \\ =2.991 s \end{gathered}$$

And 

$$\begin{gathered} t_2=\left(1.837-1.154\right)s \\ =0.683 s \end{gathered}$$

Thus, the two times for ball to elapse is 2.991 s and 0.683 s .

Step 4: (b)Determination of the horizontal and vertical components of the baseball’s velocity at each of the two times calculated in part (a).

The horizontal component of velocity is constant and unchanged because during the entire motion of the ball the acceleration is zero.

So, the horizontal component of the velocity of the baseball is given by,

$v_{0x}=v_0\cos \theta$

 

Substitute all the values in the above,

$$\begin{gathered} v_{0x}=\left(30 \mathrm{m}/\mathrm{s}\right)\cos \left(36.9°\right) \\ =24 \mathrm{m}/\mathrm{s} \end{gathered}$$

The vertical component of the velocity for time $t_1$ is given by

$v_y=v_{0y}+a_{y}t$

Here, $v_{0y}$ is the vertical component of velocity whose value is $v_0\sin \theta$, $t$ is the elapsed time, $a_y$ is the acceleration in vertical direction whose value is $-g$.

Substitute all the values in the above,

$$\begin{gathered} v_y=v_0\sin \theta -gt \\ =\left(30 \mathrm{m}/\mathrm{s}\right)\sin 36.9°-\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0.683 \mathrm{s}\right) \\ =11.3 \mathrm{m}/\mathrm{s} \end{gathered}$$

The vertical velocity at time $t_2$ is given by,

$$\begin{gathered} v_y=v_0 \sin \theta -gt \\ =\left(30 \mathrm{m}/\mathrm{s}\right)\sin 36.9°-\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(2.991 \mathrm{s}\right) \\ =-11.3 \mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the horizontal component of velocity is not change and vertical component of the velocity at $t_1$ is 11.3 m/s , and vertical component of velocity at $t_2$ is $-11.3 \mathrm{m}/\mathrm{s}$ .

Step 5: (c) Determination of the magnitude and direction of the baseball’s velocity when it returns to the level at which it left the bat

The speed with the use of projectile equations by,

$v_y^2=v_{0y}^2+a_{y}y$

Here, $v_{0y}$ is the vertical component of velocity whose value is $v_0\sin \theta$, $a_y$ is the acceleration in vertical direction whose value is $-g$ and y is the displacement of the baseball.

Substitute all the values in the above,

$$\begin{gathered} v_y^2={\left(v_0\sin \theta \right)}^2-2gy \\ ={\left[\left(30 \mathrm{m}/\mathrm{s}\right)\sin 36.9°\right]}^2-2\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(0\right) \\ =-18 \mathrm{m}/\mathrm{s} \end{gathered}$$

Here, negative sign shows the ball is falling downwards.

The magnitude of the velocity is given by,

$$\begin{gathered} v=\sqrt{v_x^2+v_y^2} \\ =\sqrt{{\left(24 \mathrm{m}/\mathrm{s}\right)}^2+{\left(-18 \mathrm{m}/\mathrm{s}\right)}^2} \\ =30 \mathrm{m}/\mathrm{s} \end{gathered}$$

The direction the velocity of the baseball is given by,

$\theta ={\tan }^{-}\left(\frac{v_y}{v_x}\right)$

Substitute all the values in the above expression,

$$\begin{gathered} \theta ={\tan }^{-}\left(\frac{-18}{24}\right) \\ =-36.9° \end{gathered}$$

Thus, the velocity of the ball when it returns is 30 m/s and the direction of the velocity of the ball is downwards in horizontal direction at the angle of $-36.9°$.