Finding Range of Possible Pitching Speeds

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    Baseball
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Homework Help Overview

The problem involves determining the pitching speed of a baseball thrown horizontally from a height of 7.00 m, landing 20.0 m away. The original poster is also considering variations in the angle of the throw, which could affect the speed calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the pitching speed using vertical and horizontal motion equations but expresses confusion regarding the impact of varying angles on the speed. Some participants suggest correcting the equations to account for the angle of throw and solving for time before substituting back into the vertical motion equation.

Discussion Status

The discussion is ongoing, with participants providing guidance on correcting the equations and exploring the implications of the angle of the throw. There is no explicit consensus yet, as participants are still working through the problem.

Contextual Notes

The original poster indicates uncertainty about the exact angle of the throw, which introduces complexity into the calculations. There may be constraints related to homework guidelines that affect how the problem can be approached.

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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 from an elevation 7.00 m above the ground. The ball lands 20.0 m away.

A. What is his pitching speed?

B. As you think about it, you’re not sure he threw the ball exactly horizontally. As you watch him throw, the pitches seem to vary from 5° below horizontal to 5° above horizontal. What is the range of speeds with which the ball might have left his hand? Enter the minimum and the maximum speed of the ball.

Homework Equations



Yf = Yo + Vit + 1/2*g*t2
Xf = Vit
Vx = Vi*cos5*t
Vy = -Vi*sin5*t

The Attempt at a Solution



A.
Yf = Yo + Vit + 1/2*g*t2
0 = 7 + 1/2*g*t2
t = 1.195s

20 = Vi(1.195s)
Vi = 16.7 m/s

B.
Y1 = Y0 + Vit-1/2*g*t2
0 = 7 + (-Vi*sin5)t - 1/2*g*t2

\Deltax = 20 =Vix*cos5*t

I'm lost at part B. Utterly.
 
Last edited:
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The horizontal equation should have v1*cos5 instead of just vi. Correct that then solve that equation for t, then plug what you get in the vertical equation. That leaves you with an equation involving vi only. Repeat with the other angle, -5.
 
kuruman said:
The horizontal equation should have v1*cos5 instead of just vi. Correct that then solve that equation for t, then plug what you get in the vertical equation. That leaves you with an equation involving vi only. Repeat with the other angle, -5.
I forgot to change it to 20 = Vix*cos5*t when I was copying it. The plugging in part was what confuses me.
 
Can you solve

20 = vix*cos5*t for t and find

t = ... ?

Can you replace t as found above in

0 = 7 + vi*sin5*t - 1/2*g*t2 ?
 

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