Home Page Textbooks Physics: Principles and Problems Page 152: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 152: Section Review

Exercise 7

Solution 1

Solution 2

Step 1

1 of 2

Since the faster ball’s air time is $textit{textbf{shorter}}$ compared to the slower ball’s, so, naturally, its $textit{textbf{vertical velocity is smaller}}$. This is also because the time the ball is exposed to gravity is shorter since it reaches the home plate faster and is therefore caught by the catcher earlier than the slower ball.

Result

2 of 2

The faster ball’s air time is $textit{textbf{shorter}}$ compared to the slower ball’s, so, naturally, its $textit{textbf{vertical velocity is smaller}}$.

Step 1

1 of 1

The faster ball stays in the air for shorter time, hence it gains smaller vertical velocity

Exercise 8

Step 1

1 of 1

Exercise 9

Step 1

1 of 2

$v_{y}^2 – v_{yi}^2 = -2 g h$

$h = dfrac{v_{y}^2 – v_{yi}^2}{-2 g}$

The y component of the velocity when the ball reaches the maximum height is zero:

$h = dfrac{(0)^2 – (v_i sin(theta))^2}{-2 g}$

$h = dfrac{(0)^2 – (11.0 sin(50))^2}{-2 (9.80)}$

$$
h = 3.62 m
$$

Result

2 of 2

$$
3.62 m
$$

Exercise 10

Solution 1

Solution 2

Step 1

1 of 2

$y = (-1/2) g t^2 + v_i sin(theta) t$

$-28 = (-1/2) (9.8) t^2 + (-15.0) sin(20) t$

$-28 = (-1/2) (9.8) t^2 – 5.1303 t$

$==> t = 1.924 s$

The horizontal displacement will be:

$$
x = v_i cos(theta) t = (15.0)*(cos20.0)*(1.924) = 27.1 m
$$

Result

2 of 2

27.1 m

Step 1

1 of 8

**Given:**
– Height: $y = 28 mathrm{~m}$;
– Velocity: $v_0 = 15 ,frac{text{m}}{text{s}}$;
– Angle: $alpha = 20°$;

**Required:**
– The horizontal distance the ball travels $s$;

Step 2

2 of 8

The ball is thrown out under the angle with initial horizontal speed and accelerates towards the ground because of the acceleration of gravity. We call that projectile motion. In projectile motion, the horizontal component of speed remains constant, and the vertical component of speed is changing due to gravity. The time of the fall is given by the second relation.
$$begin{align*}
v &= frac{s}{t} &&(1) \
{v_text y}^2 &= {v_{0, text y}}^2 + 2gy &&(2) \
v_text y – v_{0, text y} &= gt &&(3)
end{align*}$$

Step 3

3 of 8

The problem looks like this:

![‘slader’](https://slader-solution-uploads.s3.amazonaws.com/66b82e4e-e02d-48a3-90d3-fdb557c946d5-1649422377474650.png)

Step 4

4 of 8

The ball is thrown with initial velocity under the angle. That means we can rewrite it in terms of components. To do so, we use basic trigonometry identities. From the picture above, it is seen that the component in the $y$ direction is opposite to the angle $alpha$, hence we will use the definition of sine and calculate the magnitude of the component as:
$$begin{align*}
sin alpha &= frac{ v_{0, text y} }{ v_0} \
v_{0, text y} &= v_0 cdot sin alpha \
&= 15 ,frac{text{m}}{text{s}} cdot sin 20° \
&= 5.13 ,frac{text{m}}{text{s}}
end{align*}$$

Step 5

5 of 8

Now we can use the second equation to find the velocity of the ball just before it hits the ground. Taking the square root we have:
$$begin{align*}
v_text y &= sqrt{ {v_{0, text y}}^2 + 2gy } \
&= sqrt{ left( 5.13 ,frac{text{m}}{text{s}} right)^2 + 2 cdot 9.8 ,frac{text{m}}{text{s}^2} cdot 28 mathrm{~m} } \
&= 24 ,frac{text{m}}{text{s}}
end{align*}$$

Step 6

6 of 8

Dividing the third equation by $g$ we can calculate the time the tennis ball is falling:
$$begin{align*}
t &= frac{ v_text y – v_{0, text y} }{g} \
&= dfrac{ 24 ,frac{text{m}}{text{s}} – 5.13 ,frac{text{m}}{text{s}} }{ 9.8 ,frac{text{m}}{text{s}^2} } \
&= 1.92 mathrm{~s}
end{align*}$$

Step 7

7 of 8

Equal amount of time is needed for the ball to travel the vertical distance as it is needed for it to travel the horizontal distance. We calculated that time in the previous step, hence, the horizontal distance can be found multiplying the first equation by $t$ on both sides:
$$begin{align*}
s &= vt \
&= 5.13 ,frac{text{m}}{text{s}} cdot 1.92 mathrm{~s} \
&= 27.1 mathrm{~m}
end{align*}$$
$$boxed{ s = 27.1 mathrm{~m} }$$

Result

8 of 8

$$s = 27.1 mathrm{~m} $$

Exercise 11

Step 1

1 of 6

Given the scenario in which an object is thrown with the same velocity as it is thrown in Earth as it is on the Moon, where the gravity is one-sixth that of Earth’s.

Step 2

2 of 6

$textbf{a.)}$ The object’s velocity with respect to the $x$-axis, $v_x$, $textit{will not change}$ since the change in gravity does not affect the components in the $x$-axis as it is a component in the $y$ -axis.

Step 3

3 of 6

$textbf{b.)}$ The air time of the object will $textit{increase}$ since the gravitational acceleration $g$ of the object will decrease, therefore, since the relationship of $g$ to air time $t$ is inversely proportional as shown in the equation:$t = frac{-2v_y}{g}$ where $v_y$ is the velocity with respect to the $y$-axis.

Step 4

4 of 6

$textbf{c.)}$ The maximum height $y_{max}$ will also $textit{increase}$ since there is lesser gravity than on Earth, therefore there is lesser gravity pulling the object towards the ground and thus, it will reach higher compared to the situation on Earth.

Step 5

5 of 6

$textbf{d.)}$ The range $R$ of the object will also $textit{increase}$ since the horizontal velocity $v_x$ and air time $t$ also increases, such that the relationship of range $R$ to $v_x$ and $t$ is proportional as shown in the equation: $x = v_{x}t$.

Result

6 of 6

$textbf{a.)}$ $v_x$ remains $textbf{constant}$.

$textbf{b.)}$ time of flight will $textbf{increase}$.

$textbf{c.)}$ $y_{max}$ will $textbf{increase}$.

$textbf{d.)}$ $R$ will $textbf{increase}$.

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