Projectile Motion pitching speed

[charan1]

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 4.00 m above the ground. The ball lands 30.0 m away.

Part A:
What is his pitching speed?

Part 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

Sf=Si+Vis*T+.5aT^2

The Attempt at a Solution

For Part A I did the following:

Yf=Yi+Viy*T+(.5)(g)T^2

0m=4m+(0m/s)(T)+(.5)(-9.8m/s^2)(T^2)
T=.9035

Xf=Xi+Vix*T+(.5)aT^2

30m=0m+Vix(.9035 seconds)+(.5)(0m/s^2)(.9035^2)
Vix=33.204m/s


Part B:

I could only think to do the following...

cos(5)=33.2/h

h=33.3268...

I don't really know what else I can do...Please help

 

[Spinnor]

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 4.00 m above the ground. The ball lands 30.0 m away.

Part A:
What is his pitching speed?

Part 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

Sf=Si+Vis*T+.5aT^2

The Attempt at a Solution

For Part A I did the following:

Yf=Yi+Viy*T+(.5)(g)T^2

0m=4m+(0m/s)(T)+(.5)(-9.8m/s^2)(T^2)
T=.9035

Xf=Xi+Vix*T+(.5)aT^2

30m=0m+Vix(.9035 seconds)+(.5)(0m/s^2)(.9035^2)
Vix=33.204m/s


Part B:

I could only think to do the following...

cos(5)=33.2/h

h=33.3268...

I don't really know what else I can do...Please help

You need to remember the motion takes place in two dimensions. In the horizontal direction x = x_0 + v_xo*t. In the vertical direction
y = y_0 + v_yo*t + .5*a*t^2.

 

[charan1]

I don't follow. My part A is already correct.

 

[rl.bhat]

Your part A is correct.
For part B, the horizontal component of velocity will be
vix= vo*cos(theta) and vertical component of viy will be vo*sin(theta)
While calculating t, if the ball is projected up, take g -ve. And if the ball is projected down take g +ve.