How Fast Must a Baseball Travel to Clear a 29m Obstacle?

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Homework Help Overview

The problem involves a baseball being hit from a height of 1 meter at an angle of 35 degrees, aiming to clear a 29-meter obstacle located 64 meters away horizontally. Participants are tasked with finding the initial speed, time to reach the obstacle, and speed at the obstacle.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion for projectile motion, including horizontal and vertical components. There are attempts to derive relationships between the variables involved, particularly focusing on how to express time and initial speed in terms of the given distances and heights.

Discussion Status

Some participants have provided equations and suggested substituting values to find the initial speed. There is recognition of the complexity introduced by the quadratic nature of the equations, and some participants express uncertainty about their approaches. Multiple interpretations of the problem setup are being explored.

Contextual Notes

Participants note the challenge of dealing with two variables in the equations and the potential for confusion in solving the quadratic equation that arises from the projectile motion analysis.

MissJewels
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Homework Statement


A baseball is hit at a height of 1m with an initial speed of vo module at an angle of projection of 35o from the horizontal. She spent just over an obstacle of 29m high at a horizontal distance of 64m.
Find
a) vo
b) the time it takes to reach the OBSTACLES
c) its speed at the obstacle.


Homework Equations


Xi= 0
Xf= 64
Yi= 1
Yf= 29

Equations: Yf= Vosin(35) -9,8t

The Attempt at a Solution


Not too sure how to go about with this one. I end up with two variables in the equations i use. help!
 
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MissJewels said:

Homework Statement


A baseball is hit at a height of 1m with an initial speed of vo module at an angle of projection of 35o from the horizontal. She spent just over an obstacle of 29m high at a horizontal distance of 64m.
Find
a) vo
b) the time it takes to reach the OBSTACLES
c) its speed at the obstacle.
Start with:

v_x = v_0\cos(\theta)

v_y = v_0\sin(\theta) - gt

What are the equations for the height and range as a function of time?

Set height at 29 m and at horizontal range 64 m. Plug the values for x, y into these equations. From that you should be able to determine v_0.

AM
 
Andrew Mason said:
Start with:

v_x = v_0\cos(\theta)

v_y = v_0\sin(\theta) - gt

What are the equations for the height and range as a function of time?

Set height at 29 m and at horizontal range 64 m. Plug the values for x, y into these equations. From that you should be able to determine v_0.

AM

Sorry it took a while, here's what I did... something's off...
just part a)

A7543C0D.jpg
 
From the first equation:

t = 64/v_0\cos{(35)}

Substitute that value for t into the second equation:

y = y_0 + v_{0}\sin{(35)}t - \frac{1}{2}gt^2

It gets a little hairy with a quadratic equation but I think it is a little easier to solve for v0 than for t.

AM
 

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