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## Homework Statement

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 61.0 away, making a 3.00 angle with the ground.

x = 61m

Theta = 3 degrees

## Homework Equations

y = (1/2)a(t^2)

y = xtan(theta)

Vox = x/t

## The Attempt at a Solution

I first used y = xtan(theta) and got y=(61)tan(3)=3.196m From there i used y=(1/2)a(t^2) to solve for t and got t=0.808s. from there i used Vox=(61)/(0.808) and got 74.5m/s. The webste claims this is wrong. Any help?

Problem 2:

## 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 6.00 above the ground. The ball lands 30.0 away. What is his pitching speed?

I solved that part and got V=21.7m/s then there is the second part of the question.

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.

y=6m

x=30m

V=21.7m/s?

Theta=5 degree?

## Homework Equations

y = Vyt + (1/2)a(t^2)

Vy=Vtan(theta)

Vox = x/t

## The Attempt at a Solution

I honestly don't really know what to do for the second part. I tried to find the velocity for if he had thrown 5 degree above the parallel.

So i got Vy=Vtan(5)=2.4

Then i used Y=Vyt+(1/2)a(t^2)=2.4t+(-4.9)(t^2) and solved for t and got t=1.37s and then used V=x/t=30/1.37=21.9m/s

I dont know if thats right because i have yet to press submit because i can not for the life of me figure out what the velocity would be if he threw the ball 5 degrees below the parallel