Finding Range of Possible Pitching Speeds

  • Thread starter Thread starter Mirole
  • Start date Start date
  • Tags Tags
    Baseball
AI Thread Summary
To determine the pitching speed of a baseball thrown horizontally from a height of 7.00 m, the calculated speed is 16.7 m/s based on the time of flight derived from the vertical motion equation. For part B, the discussion focuses on finding the range of speeds considering a 5° variation in the throw angle. The horizontal motion equation needs to be corrected to include the cosine of the angle, leading to an equation that can be solved for time. Substituting this time into the vertical motion equation allows for the calculation of the minimum and maximum speeds. The overall approach emphasizes the importance of accurately applying trigonometric functions in projectile motion analysis.
Mirole
Messages
33
Reaction score
0

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:
Physics news on Phys.org
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 ?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top