Engineering, 28.02.2021 16:50 madisonsan9020

While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr), at an angle of 50 degrees above horizontal. The ball flies past your friend, and lands on the ground. You let the ball go at a height of 2 m above the ground. Define upward as the positive y direction, and the horizontal direction of the ball's travel as the positive x direction. You can assume that air resistance and any effects of the ball spinning are so small that they can be ignored; this means that gravity is the only force causing an acceleration. Required:
a. What is the y-component of velocity just before the ball hits the ground?
b. For how much time is the ball in the air?
c. How far horizontally does the ball travel before it hits the ground?
d. What is the magnitude of the velocity of the ball just before it hits the ground?
e. What is the angle of the total velocity of the ball just before it hits the ground?

Answers: 3

Show answers

Answers

Answer from: Lawrence101

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

- Gravitational acceleration, measured in meters per square second.

- Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball () is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground (), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Answer from: Quest

angvrswer:

vfg

grfhrtrhr:

jjj

yjujuju

jujuhuyhyt

bgf

bgr

gbr

bgr

gtr

bgr

gbgbbg

bgg

bbg

bfg

dcwf

fdb

ggb

rfe

gtg

gegte

gff

dfg

gfdsax

cvfbgnh

jmhngf

dsaazxscvb

nmhgb

fvdcsxaz

zxcv

bghn

Answer from: Quest

Idon’t do engeneering classes sorry

Another question on Engineering

Engineering, 04.07.2019 18:10

Coiled springs ought to be very strong and stiff. si3n4 is a strong, stiff material. would you select this material for a spring? explain.

Answers: 2

Answer

Engineering, 04.07.2019 18:20

Inspection for bearing condition will include: (clo4) a)-color b)-smell c)-size d)-none of the above

Answers: 1

Answer

Engineering, 04.07.2019 18:20

Avolume of 2.65 m3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 264 k, 5.6 bar. the air receives 432 kj by work from the paddle wheel. assuming the ideal gas model with cv = 0.71 kj/kg • k, determine for the air the amount of entropy produced, in kj/k

Answers: 2

Answer

Engineering, 04.07.2019 18:20

For a gate width of 2 m into the paper, determine the force required to hold the gate abc at its location.

Answers: 1

Answer

You know the right answer?

While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s...

Questions