Solve Projectile Motion: Baseball Player Hits 180m Home Run with 21.3m Height"

[erico004]

baseball player hit a home run that was estimated to have landed 180 m from home plate and to have reached a height of 21.3 m.

a. how long was it in the air?
b. what was the initial vertical velocity?
c. what was the initial horizontal velocity?
d. what was the initial velocity of the ball? ( speed and direction)

? anyone??

 

[Integral]

You need to show us something that you have tried.

You can start by finding how long it would take something to fall 21.3m.

 

[Davidrdguez]

hahah the physics homework

you have to think in the movement in two phases:

vertical and horizontal.

with the vertical one you use the formule y=vo*t+..., there are acceleration

with the horizontal one you use s=v*t, there is constant the velocity

 

[rohanprabhu]

I agree with Integral above that you should do something yourself before asking for help and as Moonbear puts it:

1) Did you show your work? Homework helpers will not assist with any questions until you've shown your own effort on the problem. Remember, we help with homework, we don't do your homework. We already passed those classes, it's your turn to do so.

However, to get you started, I am going to solve the first part for you.

i. time of ascent = time of descent

This follows by simple symmetry. It can also be proved using algebra/calculus and geometry, but i'll skip that here.

Let us consider the motion of the ball after it has reached the topmost point [initial position] along the y-axis only. From here, it falls to the ground [the final position]. At the initial position, it has a velocity = 0 [in the y-direction]. An acceleration in the y-direction = 9.8 m/s² causes it to come down.

http://img150.imageshack.us/img150/7938/projectilefbd1fs1.jpg

So, we have:
[tex]
s = -21.3~m
[/tex]
[tex]
a = -9.8~ms^{-2}
[/tex]
[tex]
u = 0
[/tex]
[tex]
t = ?
[/tex]

Using [itex] s = ut + \frac{at^2}{2}[/itex], you get

[tex]
t = 2.085~s
[/tex]

This is the time of descent. The total time will be twice of this i.e. 4.16s as it shall include the time of ascent also.

Do note that all calculations are made along the y-axis only.

 

[jake119]

yeah, rohanprabhu is right, you need to show some work first, at least make an attempt on doing it yourself before you ask for help. How are you going to pass your test if you rely on everyone else to do your work for you? We won't be there for you doing your test, so make an attempt first kid :]

 

[Distortion]

Eric, this uses the same techniques that I showed you yesterday in your horizontal velocity question.

Did my answer help you to solve that problem?