Find initial velocity of baseball given these info

  • Thread starter Libohove90
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  • #1
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Homework Statement


In a baseball game, a batter hits the ball at a height of 4.60 ft above the ground so that its angle of projection is 52.0º to the horizontal. The ball lands on the grandstand, 39.0 ft up from the bottom. The grandstand seats slope upward at 28.0º with the bottom seats 358 ft from the home plate. Calculate the speed at which the ball left the bat (ignore air resistance).

Homework Equations


(1) y - y[itex]_{}0[/itex] = v[itex]_{}0y[/itex]t - 0.5gt^2

(2) x = v[itex]_{}0x[/itex]t

The Attempt at a Solution



Since x = v[itex]_{}0x[/itex]t, we can rewrite the equation as x = v[itex]_{}0[/itex]cos[itex]\phi[/itex]t

Solve for t, we get t = x / v[itex]_{}0[/itex]cos[itex]\phi[/itex]

v[itex]_{}0y[/itex] = v[itex]_{}0[/itex]sin[itex]\phi[/itex]

Plug these variables to equation 1 and with some algebraic manipulations, you get:
y - y[itex]_{}0[/itex] = (tan [itex]\phi[/itex])x - 0.5 g( x / v[itex]_{}0[/itex]cos[itex]\phi[/itex])^2
Now I have to solve for v[itex]_{}0[/itex].

I derive this equation: v[itex]_{}0[/itex] = x[itex]\sqrt{}g / -2(cos\phi)^2(y - y_{}0 - xtan\phi[/itex])

To save you guys from more work, the given variables are y - y[itex]_{}0[/itex] = 13.7 ft
And the total distance is x = 392 ft

When I plug these variables in...I get something other than 115 ft/s which is the correct answer.

I appreciate your help.
 
Last edited:

Answers and Replies

  • #2
422
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I'm confused as to how you got the total distance to be 392ft and y-y0 to be 13.7ft.

first off, if the ball starts at y0=4.60ft and ends at y=39ft, y-y0 is most definitely not 13.7 ft, it's 34.4ft, isn't it?

Also, I got a total distance of x=431.3 ft, think about it, the slope of the grandstand is <45 degrees, so it would have to be >(358ft+39ft), which already gives you 397ft...

I think the trick to the problem is to take total x displacement and y-y0 from simple calculations (have to use some basic trig to get x) and then plug them in to your equations to find v0. You seem to have complicated it a bit...
 
  • #3
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I'm confused as to how you got the total distance to be 392ft and y-y0 to be 13.7ft.

first off, if the ball starts at y0=4.60ft and ends at y=39ft, y-y0 is most definitely not 13.7 ft, it's 34.4ft, isn't it?

Also, I got a total distance of x=431.3 ft, think about it, the slope of the grandstand is <45 degrees, so it would have to be >(358ft+39ft), which already gives you 397ft...

I think the trick to the problem is to take total x displacement and y-y0 from simple calculations (have to use some basic trig to get x) and then plug them in to your equations to find v0. You seem to have complicated it a bit...

Sorry for the confusion. Simply put, the grandstand makes 28º slope upwards with the ground. The ball lands at 39 ft from the base of the grandstand, not from the ground. The 39 ft is the hypotenuse. So in order to find y, I have to do y = 39 sin 28º = 18.3 ft. Thus y - y[itex]_{}0[/itex] = 18.3 - 4.6 = 13.7 ft. This tells me it landed 13.7 ft above the ground due to the grandstand's upward slope.

The base of the grandstand is 358 ft from the home plate. Thus to find the total distance it traveled, I have to do x = 358 + 39 cos 28º = 392 ft. So [itex]\Delta[/itex]x = 392 ft and [itex]\Delta[/itex]y = 13.7 ft. I showed my mathematical derivations and solved for v[itex]_{}0[/itex]. But when I plugged them in, I did not get 115 ft/s.

Thanks for giving the time to help me...I apologize if I made it confusing.
 
  • #4
422
4
Ohh, I see now, I assumed the 39ft was the vertical distance, not the hypotenuse, not sure why I did that, thanks for clarification. Gonna take another stab at this...
 
  • #5
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It's a great problem for those who are studying motion in two and three dimensions :)
 
  • #6
422
4
that's strange, I used your equation and values (I independently derived the same equation as you) and got the correct answer of 115ft/s.

The one thing that almost tripped me up: you're working in feet, did you remember to use the correct value of g? It's not 9.8 m/s^2, you have to convert to ft/s^2
 
  • #7
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that's strange, I used your equation and values (I independently derived the same equation as you) and got the correct answer of 115ft/s.

The one thing that almost tripped me up: you're working in feet, did you remember to use the correct value of g? It's not 9.8 m/s^2, you have to convert to ft/s^2

WOW...I cannot believe I forgot to change that. I kept everything in feet...but forgot to change g. Geez, thanks a lot :). I knew my math couldn't have been wrong. :)
 
  • #8
422
4
Yeah, like I said, that's probably the place most people would pass over. I just barely caught myself as well :P
 

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