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

[mistabry]

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 = V_iyΔt + .5g(Δt)²
Δx = V_ixΔt + 0

The Attempt at a Solution

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

Δx = 29 - 0 = V_ix * Δt
V_ix = 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?

 

[gneill]

V_iy = V sin(θ) and V_iy $$\neq$$ 0
V_ix = V cos(θ)

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

EDIT: Should have been V_iy $$\neq$$ 0. I fixed it.

 

[mistabry]

Still don't understand what you mean. Can you explain in more detail?

 

[gneill]

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.

 

[SammyS]

V_iy = V sin(θ) and V_ix $$\neq$$ 0
V_ix = V cos(θ)

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

gneill is telling you to use Vsin(θ) instead of V_iy in your y-component equation, and use Vcos(θ) instead of V_ix in your x-component equation. Use ±5° for θ.

 

[turbo21]

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.

 

[mistabry]

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

 

[noobarta]

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