Question

exam 2

1. 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.

a) how high above the ground was the baseball thrown?
b) what was the initial speed of the baseball?
c) at what height above the ground is the speed of the baseball 40 m/sec?

1. 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.

a) how fast is the arrow traveling at t = 4.00 sec? give your answer in m/sec to three significant digits.
b) what was the angle above the horizontal at which the arrow was shot? give your answer to three significant digits.

1. 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

a) the initial speed of the ball in mph to two significant digits.
b) the angle above the horizontal at which the ball was launched to two significant digits.
c) plot the vertical acceleration of the golf ball over the entire six - second flight

Explanation:

1.

a)

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

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 $v_{x}=\sqrt{v^{2}-v_{y}^{2}}$.
$v_{x}=\sqrt{50.0^{2}-39.2^{2}}=\sqrt{(50.0 + 39.2)(50.0 - 39.2)}=\sqrt{89.2\times10.8}\approx30.8\ m/s$

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:

$78.4\ m$

b)