- #1

- 26

- 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 3.00 m above the ground. The ball lands 30.0 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 degrees above horizontal. What is the range of speeds with which the ball might have left his hand?

## Homework Equations

d = vi*t + (1/2)a*t

^{2}

Trig

## The Attempt at a Solution

Problem A was completed without issue. But I'm having trouble working the possible angles into the answer.

I know that vx = vcos(5) and vy = +/- vsin(5)

I tried to isolate time, given that the horizontal distance was traveled in the same time it took the ball to hit the ground (obviously...).

vx = vcos(5) = .99v

vy = -vsin(5) = .08v

dx = vcos(5)t

30 = vcos(5)t

t = 30/(vcos(5))

3 = -vsin(5) - (1/2)(9.8)(t^2)

t = sqrt((-vsin(5) - 3)/4.9)

Now I would usually set the equations equal to each other at this point, but the result system seems too messy to be done by hand, so I feel like I'm missing a much easier solution.

Is my only option to try to find the intercept for the equations? And if the ball was thrown upwards, my equations would be the same except for the negative in from of the sin portion, correct?