How Can Pitching Speed Be Affected by Uncertainty in Throwing Angle?

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The discussion focuses on calculating a baseball player's pitching speed based on a horizontal throw from a height of 3.6 meters, resulting in a distance of 29 meters. The initial calculation yields a pitching speed of approximately 33.83 m/s. However, uncertainty in the throwing angle, ranging from 5 degrees below to 5 degrees above horizontal, complicates the determination of minimum and maximum speeds. Participants suggest using trigonometric relationships to express the vertical and horizontal components of velocity, leading to a single equation that incorporates the angle. The final equation allows for solving the pitching speed in relation to the varying angles, although some participants express confusion over the calculations and formula formatting.
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Homework Statement


A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally so that his release point is 3.6 m above the ground. The ball lands 29.0 m away.

a. What is his pitching speed?

b. As you think about it, you realize that you're not sure he threw exactly horizontally. As you watch him throw, the pitches seem to vary from 5 degrees below horizontal to 5 degrees above horizontal. Given this uncertainty in his throwing angle, what are the minimum and maximum speeds that would give result in the distance measure given above?

Homework Equations


Δy = ViyΔt + .5g(Δt)2
Δx = VixΔt + 0

The Attempt at a Solution


Δy = 0 -3.6 = 0 + .5(-9.8)(Δt)2
Δt = 6/7

Δx = 29 - 0 = Vix * Δt
Vix = 29/(6/7) = 33.83 m/s <-- Pitching speed

I got the pitching speed (part a), but I can't seem to find a way to solve part b. Any advice?
 
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Viy = V sin(θ) and Viy \neq 0
Vix = V cos(θ)

Do the math again, obtaining a single equation containing only Viy and Vix as variables (because the range, maximum x, is fixed). Substitute in the above expressions for Viy and Vix. Solve for V in terms of θ.

EDIT: Should have been Viy \neq 0. I fixed it.
 
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Still don't understand what you mean. Can you explain in more detail?
 
Take your two "Relevant equations" and form one equation where t has been eliminated (use the second equation to find an expression for t, substitute that into the first equation).

Then substitute the expressions for the x and y components of the initial velocity. You'll end up with an equation involving V and θ. Solve for V in terms of θ.
 
gneill said:
Viy = V sin(θ) and Vix \neq 0
Vix = V cos(θ)

Do the math again, obtaining a single equation containing only Viy and Vix as variables (because the range, maximum x, is fixed). Substitute in the above expressions for Viy and Vix. Solve for V in terms of θ.
gneill is telling you to use Vsin(θ) instead of Viy in your y-component equation, and use Vcos(θ) instead of Vix in your x-component equation. Use ±5° for θ.
 
omg yall are too vague! but it did give me clues on what to look for. spent like 10 hours trying to figure this problem out, but finally got it. Those in mr. todd's class don't need to thank me.

um i don't know how to post formulas cause it comes out all weird but, i found the equation in another thread here: https://www.physicsforums.com/showthread.php?t=315145
its the last post by GregD603. the equation should look something like this

y = (tanΘo)x - (g/2(Vo^2)cos^2Θo)x^2

y is vertical displacement and x horizontal displacement. theta is -5 and 5 plug everything in formula and solve for Vo and equation is also in the textbook on chap3 pg.91.
 
Lol Turbo, you're funny. This post was for DTodd's homework assignment xD! I hope you're going through the homework well! I did like all of the problems easily until I hit this one with its confusing words, but I finally got it! Aha
 
turbo21 said:
omg yall are too vague! but it did give me clues on what to look for. spent like 10 hours trying to figure this problem out, but finally got it. Those in mr. todd's class don't need to thank me.

um i don't know how to post formulas cause it comes out all weird but, i found the equation in another thread here: https://www.physicsforums.com/showthread.php?t=315145
its the last post by GregD603. the equation should look something like this

y = (tanΘo)x - (g/2(Vo^2)cos^2Θo)x^2

y is vertical displacement and x horizontal displacement. theta is -5 and 5 plug everything in formula and solve for Vo and equation is also in the textbook on chap3 pg.91.

but if you make it so that it's solving for Vo then it'll have (29tan5)-3.6 as the numerator which would make it negative and that wouldn't workout since the fraction is square rooted. Or did I do something wrong?

Do you guys have solutions to any of the other ones? all I got so far was one and five and part b of 7
 
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