exam 2\n1. a baseball is thrown horizontally from the top edge of a tall building. the ball hits the ground…

exam 2\n1. a baseball is thrown horizontally from the top edge of a tall building. the ball hits the ground moving at 50.0 meters per second 4.00 after it was thrown.\na) how high above the ground was the baseball thrown?\nb) what was the initial speed of the baseball?\nc) at what height above the ground is the speed of the baseball 40 m/sec?\n2. an arrow is shot at some initial velocity. after 4.00 seconds of flight, the arrow has a horizontal displacement of 60.0 meters and a vertical displacement of 20.0 meters.\na) how fast is the arrow traveling at t = 4.00 sec? give your answer in m/sec to three significant digits.\nb) what was the angle above the horizontal at which the arrow was shot? give your answer to three significant digits.\n3. a golf ball is hit from an elevated tee at some initial velocity. the tee is a vertical distance of 30 yards above the green and the ball lands a horizontal distance of 120 yards from the tee. if the time of flight of the ball is 6.00 seconds, calculate\na) the initial speed of the ball in mph to two significant digits.\nb) the angle above the horizontal at which the ball was launched to two significant digits.\nc) plot the vertical acceleration of the golf ball over the entire six - second flight

exam 2\n1. a baseball is thrown horizontally from the top edge of a tall building. the ball hits the ground moving at 50.0 meters per second 4.00 after it was thrown.\na) how high above the ground was the baseball thrown?\nb) what was the initial speed of the baseball?\nc) at what height above the ground is the speed of the baseball 40 m/sec?\n2. an arrow is shot at some initial velocity. after 4.00 seconds of flight, the arrow has a horizontal displacement of 60.0 meters and a vertical displacement of 20.0 meters.\na) how fast is the arrow traveling at t = 4.00 sec? give your answer in m/sec to three significant digits.\nb) what was the angle above the horizontal at which the arrow was shot? give your answer to three significant digits.\n3. a golf ball is hit from an elevated tee at some initial velocity. the tee is a vertical distance of 30 yards above the green and the ball lands a horizontal distance of 120 yards from the tee. if the time of flight of the ball is 6.00 seconds, calculate\na) the initial speed of the ball in mph to two significant digits.\nb) the angle above the horizontal at which the ball was launched to two significant digits.\nc) plot the vertical acceleration of the golf ball over the entire six - second flight

Answer

1.

a)

Explanation:

Step1: Identify vertical - motion formula

Use $y = v_{0y}t+\frac{1}{2}gt^{2}$. Since the ball is thrown horizontally, $v_{0y}=0$. $y = 0\times t+\frac{1}{2}gt^{2}$

Step2: Substitute values

Given $t = 4.00\ s$ and $g = 9.8\ m/s^{2}$, $y=\frac{1}{2}\times9.8\times4.00^{2}$ $y = 78.4\ m$

Answer:

$78.4\ m$

b)

Explanation:

Step1: Analyze horizontal and vertical components of velocity at the end

The final velocity $v = 50.0\ m/s$. First, find the vertical velocity at the end using $v_{y}=v_{0y}+gt$. Since $v_{0y} = 0$, $v_{y}=gt=9.8\times4.00 = 39.2\ m/s$.

Step2: Use Pythagorean theorem for velocity components

$v^{2}=v_{x}^{2}+v_{y}^{2}$, where $v_{x}$ is the initial horizontal speed (constant in horizontal - motion). So $v_{x}=\sqrt{v^{2}-v_{y}^{2}}$. $v_{x}=\sqrt{50.0^{2}-39.2^{2}}=\sqrt{(50.0 + 39.2)(50.0 - 39.2)}=\sqrt{89.2\times10.8}\approx30.8\ m/s$

Answer:

$30.8\ m/s$

c)

Explanation:

Step1: Let the height be $h$. First, find the vertical velocity $v_{y}$ at height $h$

$v^{2}=v_{x}^{2}+v_{y}^{2}$, we know $v = 40\ m/s$ and $v_{x}=30.8\ m/s$ (from part b)). So $v_{y}=\sqrt{v^{2}-v_{x}^{2}}=\sqrt{40^{2}-30.8^{2}}$. $v_{y}=\sqrt{(40 + 30.8)(40 - 30.8)}=\sqrt{70.8\times9.2}\approx25.6\ m/s$

Step2: Use vertical - motion formula $v_{y}^{2}=v_{0y}^{2}-2g(y - h)$

Since $v_{0y}=0$ and $y = 78.4\ m$ (from part a)), $v_{y}^{2}=- 2g(78.4 - h)$. $h = 78.4-\frac{v_{y}^{2}}{2g}=78.4-\frac{25.6^{2}}{2\times9.8}\approx44.7\ m$

Answer:

$44.7\ m$

2.

a)

Explanation:

Step1: Find horizontal and vertical components of velocity

The horizontal velocity $v_{x}=\frac{\Delta x}{t}=\frac{60.0}{4.00}=15.0\ m/s$. The vertical velocity $v_{y}=\frac{\Delta y}{t}=\frac{20.0}{4.00}=5.00\ m/s$.

Step2: Use Pythagorean theorem for the magnitude of velocity

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{15.0^{2}+5.00^{2}}=\sqrt{225 + 25}=\sqrt{250}\approx15.8\ m/s$

Answer:

$15.8\ m/s$

b)

Explanation:

Step1: Use the tangent function for the angle of projection

$\tan\theta=\frac{v_{y}}{v_{x}}$. $\theta=\arctan(\frac{5.00}{15.0})\approx18.4^{\circ}$

Answer:

$18.4^{\circ}$

3.

a)

Explanation:

Step1: Convert yards to meters

$1\ yard = 0.9144\ m$. So $120\ yards=120\times0.9144 = 109.728\ m$ and $30\ yards = 30\times0.9144 = 27.432\ m$. The horizontal velocity $v_{x}=\frac{\Delta x}{t}=\frac{109.728}{6.00}=18.288\ m/s$. In vertical - motion, $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $y=-27.432\ m$ and $t = 6.00\ s$. So $-27.432=v_{0y}\times6.00-\frac{1}{2}\times9.8\times6.00^{2}$. $v_{0y}=\frac{-27.432 + \frac{1}{2}\times9.8\times6.00^{2}}{6.00}=\frac{-27.432+176.4}{6.00}\approx24.83\ m/s$.

Step2: Find the initial speed $v_{0}$ and convert to mph

$v_{0}=\sqrt{v_{0x}^{2}+v_{0y}^{2}}=\sqrt{18.288^{2}+24.83^{2}}\approx30.8\ m/s$. To convert to mph, use $1\ m/s = 2.237\ mph$. So $v_{0}\approx30.8\times2.237\approx69\ mph$

Answer:

$69\ mph$

b)

Explanation:

Step1: Use the tangent function for the angle of projection

$\tan\theta=\frac{v_{0y}}{v_{0x}}$. $\theta=\arctan(\frac{24.83}{18.288})\approx53^{\circ}$

Answer:

$53^{\circ}$

c)

Explanation:

The vertical acceleration of the golf - ball in free - fall is constant and equal to $a=-g=- 9.8\ m/s^{2}$. To plot it, we have a horizontal axis (time $t$ from $0$ to $6.00\ s$) and a vertical axis (acceleration $a$). The plot is a horizontal line at $a=-9.8\ m/s^{2}$ for $0\leq t\leq6.00\ s$.

Answer:

The vertical acceleration is a constant $-9.8\ m/s^{2}$ for the entire $6 - s$ flight. A horizontal line at $y=-9.8$ on a graph with time on the x - axis and acceleration on the y - axis.