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

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Homework Help Overview

The problem involves a baseball being hit at an angle of 46 degrees from a height of 1.2 meters, needing to clear a 3-meter wall located 125 meters away. Participants are exploring how to determine the initial velocity required for the baseball to achieve this trajectory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the lack of horizontal acceleration and the vertical acceleration due to gravity. They explore the equations of motion in both x and y directions, questioning how to account for the wall's height and the distance to it.

Discussion Status

There are multiple lines of reasoning being explored, with participants attempting to derive equations for time and velocity. Some have provided partial equations and substitutions, while others express confusion about specific steps and calculations. Guidance has been offered to clarify the relationships between the variables involved.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also mention of additional requirements for a related project involving different angles and the creation of a chart.

jarn6700
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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.
 
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Welcome to PF!

Hi jarn6700! Welcome to PF! :smile:
jarn6700 said:
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 ahorizontal = 0. :wink:

Show us how far you get, and where you're stuck, and then we'll know how to help! :smile:
 
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 said:
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! :smile:
 
x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

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

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
 
Last edited:
jarn6700 said:
x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

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

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

Yup! :biggrin:
0 = -4.9t^2 + vsin(46)t

Nooo :cry:, it's not 0, it's 3 - 1.2 …

Try again! :smile:
 
tiny-tim said:
(try using the X2 tag just above the Reply box :wink:)

Yup! :biggrin:


Nooo :cry:, it's not 0, it's 3 - 1.2 …

Try again! :smile:


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.
 
(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 :wink:)

jarn6700 said:
3 - 1.2 = -4.9t^2 + vsin(46)t

divide both sides by t …

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

(but also why haven't you substituted the value for t from the horizontal equation, so as to give you a quadratic in vi?)
 
  • #10
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...
 
  • #11
jarn6700 said:
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 :rolleyes:)

Now solve by using the (-b ±√(b2 - 4ac))/2a formula. :smile:
 
  • #12
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
 
Last edited:
  • #13
jarn6700 said:
x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

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

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)
 
  • #14
hi mlbuxbaum! :smile:

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

from …
jarn6700 said:
x-axis 125 = Vi x cos(46)t
t = 125/Vi x cos(46)

(cos46° = 0.7) :wink:
 
  • #15
tiny-tim said:
hi mlbuxbaum! :smile:



from …


(cos46° = 0.7) :wink:

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[tex]\Theta[/tex]0-.5g(x2/v2cos2[tex]\Theta[/tex]0)
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.
 
  • #16
hi mlbuxbaum! :wink:
mlbuxbaum said:
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 :smile: was solving the two simultaneous equations …
jarn6700 said:
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 :wink:
y=y0+xtan[tex]\Theta[/tex]0-.5g(x2/v2cos2[tex]\Theta[/tex]0)
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[tex]\Theta[/tex]0-.5g(x2/v2cos2[tex]\Theta[/tex]0),

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

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