Find initial velocity given angle and distance to go over an object

[jarn6700]

Homework Statement

A baseball leaving the bat at 46 degrees at a height of 1.2m from the ground clears a 3m high wall 125 meters from home plate. what is the initial velocity of the ball?

that's my problems i am have a reasonable amount of trouble with it. any help would be much appreciated.

 

[tiny-tim]

Welcome to PF!

Hi jarn6700! Welcome to PF!

A baseball leaving the bat at 46 degrees at a height of 1.2m from the ground clears a 3m high wall 125 meters from home plate. what is the initial velocity of the ball?

that's my problems i am have a reasonable amount of trouble with it. any help would be much appreciated.

ok … use the usual constant acceleration equations, for the x and y directions separately, and remember that there is no horizontal force, so a_horizontal = 0.

Show us how far you get, and where you're stuck, and then we'll know how to help!

 

[jarn6700]

yeah i gather there is no acceleration in the x-axis and that there is the
-9.81 acceleration in the y-axis. but i can't seem to figure out firstly the velocity to reach the 125m and then secondly how to account for the 3m high wall.

i've had a read of some general initial velocity thread which were a bit confusing (i.e. https://www.physicsforums.com/showthread.php?t=314614&highlight=initial+velocity ) so anything to start us off is great.

cheers

 

[jarn6700]

this one too https://www.physicsforums.com/showthread.php?t=184043 but i don't get how they got to their conclusions.

 

[tiny-tim]

yeah i gather there is no acceleration in the x-axis and that there is the
-9.81 acceleration in the y-axis. but i can't seem to figure out firstly the velocity to reach the 125m and then secondly how to account for the 3m high wall.

x goes from 0 to 125, and y goes from 0 to 3 …

call the total time "t" …

that should give you two equations, and then you can eliminate t.

Show us how far you get, and where you're stuck, and then we'll know how to help!

 

[jarn6700]

x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

then do i sub that into
<br /> y=-\frac{1}{2}g t^2+v_{y,0}t ?

0 = -4.9t^2 + vsin(46)t

divide both sides by t getting

0 = -4.9t + vsin(46)

4.9t = vsin(46)

4.9 = vsin(46)/t

4.9 = 0.72v/125/0.7v

 

[tiny-tim]

x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

then do i sub that into
<br /> y=-\frac{1}{2}g t^2+v_{y,0}t ?

(try using the X² tag just above the Reply box )

Yup!

0 = -4.9t^2 + vsin(46)t

Nooo , it's not 0, it's 3 - 1.2 …

Try again!

 

[jarn6700]

(try using the X² tag just above the Reply box )

Yup!


Nooo , it's not 0, it's 3 - 1.2 …

Try again!

not sure what you are referring to with the x^2 tag (please explain :) )


3 - 1.2 = -4.9t^2 + vsin(46)t

divide both sides by t getting

1.8/t = -4.9t + vsin(46)

1.8/t + 4.9t = vsin(46)

and now i am stuck. yup, very stuck.

 

[tiny-tim]

(if you click the QUOTE button, you get the Reply page, and just above the text field there's a line that begins B I U … there's an X² and X₂ tag near the end, and if you click them it gives you supersripts and subscripts )

3 - 1.2 = -4.9t^2 + vsin(46)t

divide both sides by t …

why?? … this is a straightforward quadratic equation as it is!

(but also why haven't you substituted the value for t from the horizontal equation, so as to give you a quadratic in v_i?)

 

[jarn6700]

1.8 = -4.9(125/0.7v)^2 + 0.72v(125/0.7v)

0 = -875v^2 + 128.57v -1.8is that correctly put into the equation? sorry i battle with quadratics...

 

[tiny-tim]

1.8 = -4.9(125/0.7v)^2 + 0.72v(125/0.7v)

0 = -875v^2 + 128.57v -1.8


is that correctly put into the equation? sorry i battle with quadratics...

Yes (except you didn't square the 125 )

Now solve by using the (-b ±√(b² - 4ac))/2a formula.

 

[jarn6700]

whe you say i didnt square the 125: i thought you were meant to leave the (125/0.7v)^2 as a squared number to be able to put into the formula?

-76564.5v + 128.7v - 1.8

(-128.7 ±√(128.7)^2 - 4 (-76564.5)(-1.8))/2(-76564.5)
(-128.7 ±√534700.71)/-153129
(-128.7 ±731.23)/-153129
that right so far?

(did -875v^2 + 128.57v -1.8 just incase, wasnt sure.)
(-128.7 ±√10263.69)/-1750
(-128.7 ± 101.31)/-1750

thanks for your help so far, been very useful

 

[mlbuxbaum]

x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

then do i sub that into
<br /> y=-\frac{1}{2}g t^2+v_{y,0}t ?

0 = -4.9t^2 + vsin(46)t

divide both sides by t getting

0 = -4.9t + vsin(46)

4.9t = vsin(46)

4.9 = vsin(46)/t

4.9 = 0.72v/125/0.7v

can someone tell me where 0.7v in the last line came from? I get that .72v=vsin(46) and t=125 but i am lost as to where that 0.7v snuck in...it also took me a while to figure out that the x value is time. I thought of it as distance not time. I was lost, but i am now found(sort of)

 

[tiny-tim]

hi mlbuxbaum!

can someone tell me where 0.7v in the last line came from?

from …

x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

(cos46° = 0.7)

 

[mlbuxbaum]

hi mlbuxbaum!



from …


(cos46° = 0.7)

I figured that .72 was sin46°(actually .7193), but in the eq he shows
4.9=vsin(46)/t
4.9=.72v/125/.7v

not sure why the equation is divided by cos46°.

my problem is alittle more meaty than this original.(to me)
This is for my Intro to Engineering class, for the Engineering with Excel portion. The instructor wants us to find initial velocity at 15,30,45 & 60°, create a chart(graph) for each angle by creating a chart of 11 x, y coordinates for each angle. He gave us a formula to use,
y=y₀+xtan\Theta₀-.5g(x²/v²cos²\Theta₀)
Then he told us to set y=0 to get the range, and then use the quadratic equation to get the x.

I really like how you help walk us through these problems. Just the answers don't help , i want to learn how to do it. I really hope you can help me with this. I am most likely over-thinking this and losing myself in the process.

 

[tiny-tim]

hi mlbuxbaum!

I figured that .72 was sin46°(actually .7193), but in the eq he shows
4.9=vsin(46)/t
4.9=.72v/125/.7v

not sure why the equation is divided by cos46°.

jarn6700 was solving the two simultaneous equations …

t = 125/Vi x cos(46)
0 = -4.9t^2 + vsin(46)t

…

"simultaneous" means that they have the same value of t,

so we can eliminate t by solving for t in one equation, then substituting that value of t in the other equation

the cos46° comes from the first equation

y=y₀+xtan\Theta₀-.5g(x²/v²cos²\Theta₀)
Then he told us to set y=0 to get the range, and then use the quadratic equation to get the x.

i don't see the difficulty …

you have 0 = y₀+xtan\Theta₀-.5g(x²/v²cos²\Theta₀),

you put in one given angle at a time, and use the standard formula for solving a quadratic equation (-b ±√(b² - 4ac))