Solving a Projectile Motion Problem: Maximum Height of a Batted Baseball

[raizen91]

hello guys! i find physics really difficult and my classmates and i have been dumbfounded by this problem. hope you could help me with the formulas i should use and how is should use them. this is was the first problem: ooops..this problem aint a trajectory problem. its a projectile motion problem. pls help, i don't know what our teacher expect from us 10 year olds grade school students...sigh

a baseball is batted into the air and caught at a point 100m distant horizontally in 4 seconds. if air resistance is neglected, what is its maximum height in meters above the ground? thnx

 

[Harmony]

May I know if the angle of the projectile motion is given?

 

[GregA]

You are looking for the distance an object will travel such that the vertical component of it's velocity = 0 at a certain point in time

Do you know at what time the ball will reach it's greatest height?

Do you know what force will cause the ball to achieve a (vertical) speed of 0m/s from v m/s?

 

[vaishakh]

The given information is certianly insufficient.

 

[GregA]

I disagree...given that the ball is caught at the same height as it was thrown (one would reasonably assume) we can infer the time at which it is at its greatest height..if we really wanted to we could also find its speed because wecan find not only the vertical component of its initial velocity but also the horizontal too...and finally with this we can find the angle of projection.

All that is needed though is the distance the ball will travel before it starts to fall back down again.

 

[vaishakh]

Sorry. I misread the question. Okay to assume that ball is caught at the same height at which it is shot is something which need not be said. However I didn't see the 4 seconds time given.
So you know horizontal component as well as vertical component and can certianly calculate the initial velocit as well as projection anlge.
Very Very Sorry.

 

[Harmony]

Is the answer 125m? (assuming g=10ms^-2)
1. Horizantal velocity = horizantal distance / time
2. Vertical velocity =( horizantal velocity / cos x (x=angle) ) * sin x
3. sinx/cosx=tan x (trigonometry)
3. Vertical velocity = horizantal velocity * tan x
4. Since v^2=u^2-2gh, v=0,
5. (vertical velocity)^2=2gh
6. To get the highest value of h, x must be 45 degrees. (tan45=1)
7. Calculate the height now.

 

[tony873004]

10 years old? You were given this problem in the 3rd or 4th grade?

 

[raizen91]

sorry...that's all the info our teacher gave us. my classmates and i had the same prob... like no angle?

 

[raizen91]

yap I am ten..but I am in my 5th grade. i find it quite silly when our teacher thinks that we're smart enough to answer questions as difficult as this..

 

[raizen91]

let me try that harmony...hmmm...tnx

 

[andrevdh]

The launch angle $\theta_o$ can be calculated via the 4s for an object to fall from rest downwards to the point where the ball is catched
"[PLAIN Forums/projectile launch angle.htm"]http://mysite.mweb.co.za/residents/andriesvdh/Physics Forums/projectile launch angle.htm[/URL]

Since $v_x=25\ m/s$
it follows that the initial upwards velocity is given by
$v_{yo}=25\times0.784$

 

[HallsofIvy]

Actually, there is more information than necessary! That "100 m" is not necessary. It's also not necessary to worry about the angle. You have two separate equations for the horizontal and vertical distances.

horizontal: x= v_xt
vertical: y= -4.9t²+ v_yt

Since the ball moves 100 m horizontally, in 4 seconds, we have
100= 4v_x and 0= -4.9(16)+ 4v_y.

We really only need the second equation: 78.4+ 4v_y= 0 can be solved for v_y. After you know that you can find the maximum height by completing the square in y= -4.9t²+ v_yt.

If the ball were hit straight up and came back down to the same level in 4 seconds or hit so that it went 1000m horizontally while coming back to the same level in 4 seconds, it would still have the same maximum height.

 

[raizen91]

Actually, there is more information than necessary! That "100 m" is not necessary. It's also not necessary to worry about the angle. You have two separate equations for the horizontal and vertical distances.

horizontal: x= v_xt
vertical: y= -4.9t²+ v_yt

Since the ball moves 100 m horizontally, in 4 seconds, we have
100= 4v_x and 0= -4.9(16)+ 4v_y.

We really only need the second equation: 78.4+ 4v_y= 0 can be solved for v_y. After you know that you can find the maximum height by completing the square in y= -4.9t²+ v_yt.

If the ball were hit straight up and came back down to the same level in 4 seconds or hit so that it went 1000m horizontally while coming back to the same level in 4 seconds, it would still have the same maximum height.

will the answer be 19.6? i don't know about ur formula but i have used a different approach. since the the time is 4 sec., i would assume that the ball will reach its highest point at 2 sec. so i used two formulas which is
v^2 = v_{0^2+ 2as and
v = v0 + at
so
0 = vo + (-9.8)(2)
vo = 19.6 m/s
0= (19.6)^{2 + 2 (-9.8)s
-(19.6)2/-19.6 = -19.6s/-19.6
s= 19.6
this is my final answer..19.6 being the displacement}}

 

[Haxx0rm4ster]

lol he is either 15 or 14.
Inferring that the 91 in his name is the year when he was born.

 

[andrevdh]

Yes raizen you've got it! Note that my $v_{yo}$ comes to the same as yours.

 

[raizen91]

:tongue2: ......

 

[raizen91]

tnx for the help!

 

[andrevdh]

It's a pleasure.

 

[raizen91]

i got two more problems, hope you could help. more projectile problems...(sigh), i don't know how to start, hoping you guys could help me with the initial steps...1. a missile is fired with a launch velocity of 15,000 ft/s at a target 1,200 miles away. at what angle must it be fired to hit the target? how long after it is fired will the target be hit.2. a projectile is fired at an angle of 30 degrees above the horizontal from the top of a cliff 600 ft. high : the initial speed of the projectile is 2000 ft/s. how far will the projectile move horizontally before it hits level grained at the base of the cliff?

 

[andrevdh]

1. Using for the x-velocity component of the projectile
$v_x=v_o\cos(\theta_o)$
you can obtain an equation for the flight time. Use this time
in the equation for $y=o$

2. For this the y velocity component is given by
$v_y=v_o\sin(\theta_o)$
Use this and the fact that when it lands
$y=-600$

 

[raizen91]

1. Using for the x-velocity component of the projectile
$v_x=v_o\cos(\theta_o)$
you can obtain an equation for the flight time. Use this time
in the equation for $y=o$

2. For this the y velocity component is given by
$v_y=v_o\sin(\theta_o)$
Use this and the fact that when it lands
$y=-600$

ok, what is did was to first convert ft and miles to meters. using the formula you gave v(sub)x = v(sub)o cos (thetha)
1931212.8(1200 miles) = 4572cos(theta)

then for y
0 = 4572sin(theta)*t - 4.9<sup>2

err..ok. am i at the right path?</sup>

 

[andrevdh]

The first equation describes the x velocity component of the motion of the projectile. Since the x velocity of the projecile is constant one can say that
$$v_x=r/t$$
where $r$ is the range of the projectile and $t$ is the flight time. Use this flight time for the second equation, which I think you intented to include the $t^2$ term.

There are usually many ways to solve these projectile problems. So be warned someone may come up with a more elegant solution, but what is important is that you understand what you are doing now. So if you have any questions about what is happening up to this point feel free to ask.

 

[alfaphysics]

hello guys! i find physics really difficult and my classmates and i have been dumbfounded by this problem. hope you could help me with the formulas i should use and how is should use them. this is was the first problem: ooops..this problem aint a trajectory problem. its a projectile motion problem. pls help, i don't know what our teacher expect from us 10 year olds grade school students...sigh

a baseball is batted into the air and caught at a point 100m distant horizontally in 4 seconds. if air resistance is neglected, what is its maximum height in meters above the ground? thnx

Horizontal distance = 100m
time 4 seconds
Horizontal speed = 25m/s
Horizontal range = 2 [h velocity][vertical velocity]/g
vertical velocity = Rg/2[horizontal velocity]= 1000/50 = 20
maximum height = 20 x 20/20= 20m

 

[raizen91]

err...

The first equation describes the x velocity component of the motion of the projectile. Since the x velocity of the projecile is constant one can say that
$$v_x=r/t$$
where $r$ is the range of the projectile and $t$ is the flight time. Use this flight time for the second equation, which I think you intented to include the $t^2$ term.

There are usually many ways to solve these projectile problems. So be warned someone may come up with a more elegant solution, but what is important is that you understand what you are doing now. So if you have any questions about what is happening up to this point feel free to ask.

thanks for the help but I am really confused, can you walk me through your solution. you're right, I've been provided with other solutions but they only made me more confused. id like to stick to how you do it. can you to start from the top? and can you explain using simple english? because my english is not that good... thnx

 

[raizen91]

The first equation describes the x velocity component of the motion of the projectile. Since the x velocity of the projecile is constant one can say that
$$v_x=r/t$$
where $r$ is the range of the projectile and $t$ is the flight time. Use this flight time for the second equation, which I think you intented to include the $t^2$ term.

There are usually many ways to solve these projectile problems. So be warned someone may come up with a more elegant solution, but what is important is that you understand what you are doing now. So if you have any questions about what is happening up to this point feel free to ask.

ok, here's what i did.
v_{x = r / t
dx/ v[subx] = vx * t / vx
= 1931,216.66 m / 4572.009
= 422.4 seconds
now for the angle
cos[theta] = vx / vx
= 4572.009 / 4572.009 =1
= 60 degrees
ok...is this right?}

 

[dontcare]

For the second projectile problem. You need to find the x and y vector components in order to solve for the amount of time. After you calculate the time you will be able to solve for the horizontal distance.

$$v_ox = v_o cos\theta$$
$$v_oy = v_o sin\theta$$

$$\Delta y = v_oy t - \frac{1}{2}gt^2$$
$$\Delta x = v_o cos\theta * t$$

 

[andrevdh]

ok, here's what i did.
v_{x = r / t
dx/ v[subx] = vx * t / vx
= 1931,216.66 m / 4572.009
= 422.4 seconds}

_{you cannot calculate $v_x$ yet since you don't have $\theta_o$ - the x-velocity component is given by $v_x=v_o\cos(\theta_o)$. You have used $v_o$ in your equation.}

 

[andrevdh]

Your previous attempt:

ok, what is did was to first convert ft and miles to meters. using the formula you gave v(sub)x = v(sub)o cos (thetha)
1931212.8(1200 miles) = 4572cos(theta)

then for y
0 = 4572sin(theta)*t - 4.9<sup>2

err..ok. am i at the right path?</sup>

<sup>is better - you are calculating the x velocity component with the first equation, so the range $r=1.93\times 10^6$ needs to be divided by the time of flight $t_f$ of the projectile to get the x-velocity component.

Then solve this equation for $t_f$ and substitute it into the second equation, which needs a $t^2$ in the last part.</sup>