part a: concept check (15 points)
1. what is the acceleration due to gravity near the earth’s surface? state its standard value and direction.
2. how does velocity change during upward motion compared to downward motion in free - fall?
3. in projectile motion (without air resistance), which velocity component remains constant and which changes? explain why.
4. if two objects are dropped at the same time, one vertically downward and the other launched horizontally, which hits the ground first? why?
5. a student claims, “the horizontal motion affects the vertical motion in projectile motion.” do you agree or disagree? explain.
part b: guided exploration (20 points)
1. a ball is dropped from a height of 45 m (ignore air resistance).
a. how long does it take to reach the ground?
b. what is its velocity just before hitting the ground?
2. a rock is thrown straight upward with an initial velocity of 20 m/s.
a. how high does it rise?
b. how long does it take to return to the thrower’s hand?
part c: application / problem solving (35 points)
show complete solutions.
1. a basketball player throws a ball at 12 m/s at an angle of 50° above the horizontal.
a. how long is the ball in the air?
b. what is the maximum height reached?
c. how far does the ball travel horizontally before hitting the ground?
2. a stone is launched horizontally from the top of a 78.4 m cliff with an initial velocity of 15 m/s.
a. how long does it take to reach the ground?
b. how far from the base of the cliff does it land?

Question

part a: concept check (15 points)
1. what is the acceleration due to gravity near the earth’s surface? state its standard value and direction.
2. how does velocity change during upward motion compared to downward motion in free - fall?
3. in projectile motion (without air resistance), which velocity component remains constant and which changes? explain why.
4. if two objects are dropped at the same time, one vertically downward and the other launched horizontally, which hits the ground first? why?
5. a student claims, “the horizontal motion affects the vertical motion in projectile motion.” do you agree or disagree? explain.
part b: guided exploration (20 points)
1. a ball is dropped from a height of 45 m (ignore air resistance).
 a. how long does it take to reach the ground?
 b. what is its velocity just before hitting the ground?
2. a rock is thrown straight upward with an initial velocity of 20 m/s.
 a. how high does it rise?
 b. how long does it take to return to the thrower’s hand?
part c: application / problem solving (35 points)
show complete solutions.
1. a basketball player throws a ball at 12 m/s at an angle of 50° above the horizontal.
 a. how long is the ball in the air?
 b. what is the maximum height reached?
 c. how far does the ball travel horizontally before hitting the ground?
2. a stone is launched horizontally from the top of a 78.4 m cliff with an initial velocity of 15 m/s.
 a. how long does it take to reach the ground?
 b. how far from the base of the cliff does it land?

Accepted Answer

### Part A
1. **Brief Explanation**: The acceleration due to gravity near Earth's surface is approximately $9.8\ m/s^{2}$ and its direction is downward towards the center of the Earth.
**Answer**: The standard value of acceleration due to gravity near Earth's surface is $g = 9.8\ m/s^{2}$ and its direction is downward.
2. **Brief Explanation**: In free - fall, during upward motion, velocity decreases due to the downward pull of gravity. During downward motion, velocity increases as gravity acts in the same direction as the motion.
**Answer**: During upward motion in free - fall, velocity decreases; during downward motion, velocity increases because gravity acts downward.
3. **Brief Explanation**: In projectile motion without air resistance, the horizontal velocity component remains constant because there is no horizontal acceleration. The vertical velocity component changes due to the acceleration of gravity.
**Answer**: The horizontal velocity component remains constant and the vertical velocity component changes because there is no horizontal acceleration and there is a vertical acceleration due to gravity.
4. **Brief Explanation**: Both objects will hit the ground at the same time. The vertical motion of both objects is a free - fall motion with the same initial vertical velocity (0 for the dropped object and 0 in the vertical direction for the horizontally launched object) and the same acceleration due to gravity.
**Answer**: Both objects hit the ground at the same time because their vertical motions are identical free - fall motions.
5. **Brief Explanation**: The horizontal and vertical motions in projectile motion are independent of each other. The horizontal motion has a constant velocity and the vertical motion is a free - fall motion.
**Answer**: Disagree. Horizontal and vertical motions in projectile motion are independent.

### Part B
1.
a. **Explanation**:
## Step 1: Use the equation $h=v_0t+\frac{1}{2}gt^{2}$, where $h = 45\ m$, $v_0 = 0\ m/s$ (dropped), and $g=9.8\ m/s^{2}$.
The equation simplifies to $h=\frac{1}{2}gt^{2}$. Solving for $t$, we get $t=\sqrt{\frac{2h}{g}}$.
## Step 2: Substitute $h = 45\ m$ and $g = 9.8\ m/s^{2}$ into the formula.
$t=\sqrt{\frac{2	imes45}{9.8}}\approx\sqrt{\frac{90}{9.8}}\approx 3.03\ s$
**Answer**: $t\approx3.03\ s$
b. **Explanation**:
## Step 1: Use the equation $v = v_0+gt$, where $v_0 = 0\ m/s$, $g = 9.8\ m/s^{2}$, and $t\approx3.03\ s$ (from part a).
$v=0 + 9.8	imes3.03=29.69\ m/s$
**Answer**: $v\approx29.69\ m/s$
2.
a. **Explanation**:
## Step 1: At the maximum - height, the final vertical velocity $v = 0\ m/s$. Use the equation $v^{2}-v_0^{2}=2ah$, where $v_0 = 20\ m/s$, $v = 0\ m/s$, and $a=-g=- 9.8\ m/s^{2}$.
Solving for $h$, we get $h=\frac{v^{2}-v_0^{2}}{2a}=\frac{0 - 20^{2}}{2	imes(-9.8)}$.
## Step 2: Calculate the value of $h$.
$h=\frac{-400}{-19.6}\approx20.41\ m$
**Answer**: $h\approx20.41\ m$
b. **Explanation**:
## Step 1: Use the equation $v = v_0+at$. When the rock returns to the thrower's hand, the displacement $y = 0$. We can also use the equation $v = v_0 - gt$. At the end of the motion, when it returns to the starting point, if we consider the initial velocity $v_0 = 20\ m/s$ and $a=-g=-9.8\ m/s^{2}$, and the final velocity $v=-20\ m/s$ (same speed but opposite direction).
Using $v = v_0+at$, we solve for $t$: $t=\frac{v - v_0}{a}=\frac{-20 - 20}{-9.8}=\frac{-40}{-9.8}\approx4.08\ s$
**Answer**: $t\approx4.08\ s$

### Part C
1.
a. **Explanation**:
## Step 1: First, find the initial vertical velocity $v_{0y}=v_0\sin	heta$, where $v_0 = 12\ m/s$ and $	heta = 50^{\circ}$. So $v_{0y}=12\sin50^{\circ}\approx9.19\ m/s$.
The time of flight $t$ can be found using the equation $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$. When the ball returns to the same height ($y - y_0 = 0$), $0=v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (initial time), and the other is $t=\frac{2v_{0y}}{g}$.
## Step 2: Substitute $v_{0y}\approx9.19\ m/s$ and $g = 9.8\ m/s^{2}$ into the formula.
$t=\frac{2	imes9.19}{9.8}\approx1.88\ s$
**Answer**: $t\approx1.88\ s$
b. **Explanation**:
## Step 1: At the maximum height, the vertical velocity $v_y = 0$. Use the equation $v_y^{2}-v_{0y}^{2}=2ah$, where $a=-g=-9.8\ m/s^{2}$, $v_y = 0$, and $v_{0y}\approx9.19\ m/s$.
Solving for $h$, we get $h=\frac{v_y^{2}-v_{0y}^{2}}{2a}=\frac{0 - 9.19^{2}}{2	imes(-9.8)}$.
## Step 2: Calculate the value of $h$.
$h=\frac{-84.46}{-19.6}\approx4.31\ m$
**Answer**: $h\approx4.31\ m$
c. **Explanation**:
## Step 1: First, find the initial horizontal velocity $v_{0x}=v_0\cos	heta$, where $v_0 = 12\ m/s$ and $	heta = 50^{\circ}$. So $v_{0x}=12\cos50^{\circ}\approx7.71\ m/s$.
The horizontal distance $x$ is given by $x = v_{0x}t$, where $t\approx1.88\ s$ (from part a).
## Step 2: Calculate $x$.
$x=7.71	imes1.88 = 14.49\ m$
**Answer**: $x\approx14.49\ m$
2.
a. **Explanation**:
## Step 1: The initial vertical velocity $v_{0y}=0\ m/s$. Use the equation $y - y_0=v_{0y}t+\frac{1}{2}gt^{2}$, where $y - y_0=-78.4\ m$ (downward), $v_{0y}=0\ m/s$, and $g = 9.8\ m/s^{2}$.
The equation simplifies to $y - y_0=\frac{1}{2}gt^{2}$. Solving for $t$, we get $t=\sqrt{\frac{2|y - y_0|}{g}}$.
## Step 2: Substitute $|y - y_0| = 78.4\ m$ and $g = 9.8\ m/s^{2}$ into the formula.
$t=\sqrt{\frac{2	imes78.4}{9.8}}=\sqrt{16}=4\ s$
**Answer**: $t = 4\ s$
b. **Explanation**:
## Step 1: The initial horizontal velocity $v_{0x}=15\ m/s$. The horizontal distance $x$ is given by $x = v_{0x}t$, where $t = 4\ s$ (from part a).
$x=15	imes4=60\ m$
**Answer**: $x = 60\ m$

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Question

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Part A

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