A ball is hit with initial velocity -- Find the Horiz/Vert displacement

[Rijad Hadzic]

Homework Statement

A softball is hit with an initial velocity of 29 m/s. at an angle of 60 degrees above the horizontal and impacts the top of the outfield fence 5s later. Assuming the initial height of the softball was .5 m above (level) ground, what are the balls horizontal and vertical displacements?

Homework Equations

$a_xt + V_i = V_f$
$\Delta x = (1/2) (V_f + V_i ) t$
$\Delta x = V_it + (1/2)a_xt^2$
$\Delta x = V_f - (1/2) a_xt^2$[/B]

The Attempt at a Solution

[itex] V_y = 29sin(60) = 25.11473671 m/s[itex]
$V_x = 29cos(60) = 14. 5 m/s$
Horizontal displacement is easy. Using formula $\Delta x = (1/2) (V_f + V_i ) t$ with $V_f = V_i$ you get $\Delta x = V_ft$, so horizontal displacement = (14.5 m/s)(5s) = 72.5 m

Now to find vertical displacement:

I use formula $a_xt + V_i = V_f$ to find final velocity, plugging in I get:
$(-9.8 m/s^2) (5s) + 25.11473671 m/s = V_f$
= $-23.88526329 m/s = V_f$

Now using $\Delta y = (1/2)(V_f + V_i)(t)$ I get:

$\Delta y = (1/2)(-23.88526329 m/s + 25.11473671 m/s) (5s)$

= 3.07368355 m = y displacement

But the softball starts off with in initial height of .5 m, so I subtract y displacement - .5, which =

2.57368355 m,

but my book gives me answer: 2.95 m

Can anyone tell me where I went wrong here?? I don't see my mistake..

 

[gneill]

You're taking several differences of numbers that are fairly close in magnitude. You are keeping lots of extra digits to maintain accuracy, which is good, however it doesn't make sense to use such accuracy if one of your starting values has only one decimal place of accuracy. I'm referring to the value of g that you're using.

Try using a more accurate value for g, such as $g = 9.80665~m/s^2$.

Alternatively do all the work symbolically first, only plugging in numbers when you've reduced the expression and removed all redundant operations. It's a less error prone way to work, avoiding rounding and truncation issues as well as transcription errors from carrying along all those digits through the algebra.

 

[Rijad Hadzic]

You're taking several differences of numbers that are fairly close in magnitude. You are keeping lots of extra digits to maintain accuracy, which is good, however it doesn't make sense to use such accuracy if one of your starting values has only one decimal place of accuracy. I'm referring to the value of g that you're using.

Try using a more accurate value for g, such as $g = 9.80665~m/s^2$.

Alternatively do all the work symbolically first, only plugging in numbers when you've reduced the expression and removed all redundant operations. It's a less error prone way to work, avoiding rounding and truncation issues as well as transcription errors from carrying along all those digits through the algebra.

I have tried using that value of g and there was an even bigger difference from the books answer to the one I got...

I'm not really sure where the error has come from after I've done it algebraically as well... Does my method look good to you?

 

[PeroK]

I have tried using that value of g and there was an even bigger difference from the books answer to the one I got...

I'm not really sure where the error has come from after I've done it algebraically as well... Does my method look good to you?

The $\Delta y$ you calculated is the y-displacement. There is no need to subtract $0.5m$.

However, to echo what @gneill says, your use of numbers to umpteen decimal places is clumsy, makes your work difficult to read and even more difficult to spot an error, and is technically wrong (given you do not know the input data to this degree of accuracy).

Finally, it would have been simpler to get both displacements from:

$\Delta x = v_x t$ and $\Delta y = v_yt - \frac12 gt^2$

 

[TomHart]

I think the book used g = 9.81 m/s^2, instead of 9.8 m/s^2.

 

[Rijad Hadzic]

The $\Delta y$ you calculated is the y-displacement. There is no need to subtract $0.5m$.

However, to echo what @gneill says, your use of numbers to umpteen decimal places is clumsy, makes your work difficult to read and even more difficult to spot an error, and is technically wrong (given you do not know the input data to this degree of accuracy).

Finally, it would have been simpler to get both displacements from:

$\Delta x = v_x t$ and $\Delta y = v_yt - \frac12 gt^2$

Why is there no need to subtract .5 m? Isn't the formula $\Delta y = V_yt - \frac 12 gt^2$ assuming that it starts at level with the ground?

 

[TomHart]

Isn't the formula Δy=V_yt−1/2gt² assuming that it starts at level with the ground?

Δy means the change in y, or the displacement in the y direction. The displacement will always be the same, regardless of the starting position. The problem statement asked for the horizontal and vertical displacements - not position.

 

[PeroK]

Why is there no need to subtract .5 m? Isn't the formula $\Delta y = V_yt - \frac 12 gt^2$ assuming that it starts at level with the ground?

No, that equation makes no assumption about a starting height, because it calculates $\Delta y$.

In fact, it would have made more sense to add the $0.5m$, which would have given you the height of the ball.

 

[Rijad Hadzic]

Ahh alright. I got you guys. I understand now, displacement formula = displacement, lol. Doesn't matter where it starts.

I got pretty close to the answer, with the answer being 2.94868355

Apparently I need to use the full value for initial y velocity (25.11473671) and g = 9.81

 

[PeroK]

Ahh alright. I got you guys. I understand now, displacement formula = displacement, lol. Doesn't matter where it starts.

I got pretty close to the answer, with the answer being 2.94868355

Apparently I need to use the full value for initial y velocity (25.11473671) and g = 9.81

Do you think you can calculate the position of a softball to the nearest nanometer?

 

[Rijad Hadzic]

Do you think you can calculate the position of a softball to the nearest nanometer?

No, so my answer would have to be rounded to 2.95 meters! Which is the same as my books answer.