part a: concept check (15 points)\n1. what is the acceleration due to gravity near the earth’s surface…

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

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.

$t=\sqrt{\frac{2\times45}{9.8}}\approx\sqrt{\frac{90}{9.8}}\approx 3.03\ s$ 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).

$v=0 + 9.8\times3.03=29.69\ m/s$ 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 $h=\frac{v^{2}-v_0^{2}}{2a}=\frac{0 - 20^{2}}{2\times(-9.8)}$.

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

a. Explanation:

Step 1: First, find the initial vertical velocity $v_{0y}=v_0\sin\theta$, where $v_0 = 12\ m/s$ and $\theta = 50^{\circ}$. 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.

$t=\frac{2\times9.19}{9.8}\approx1.88\ s$ 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 $h=\frac{v_y^{2}-v_{0y}^{2}}{2a}=\frac{0 - 9.19^{2}}{2\times(-9.8)}$.

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 $v_{0x}=v_0\cos\theta$, where $v_0 = 12\ m/s$ and $\theta = 50^{\circ}$. 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 $x$.

$x=7.71\times1.88 = 14.49\ m$ 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.

$t=\sqrt{\frac{2\times78.4}{9.8}}=\sqrt{16}=4\ s$ 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).

$x=15\times4=60\ m$ Answer: $x = 60\ m$