Finding Range of Possible Pitching Speeds

[Mirole]

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

$$\Delta$$x = 20 =Vix*cos5*t

I'm lost at part B. Utterly.

 

[kuruman]

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.

 

[Mirole]

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.

 

[kuruman]

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 ?