Problem with projectile velocity

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Homework Help Overview

The problem involves a ball thrown from a building at a certain height, with specific parameters regarding its motion, including time of flight, horizontal distance, and impact angle. The goal is to determine the height of the building and the initial velocity components of the ball.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using trigonometric relationships to find the height based on the angle and horizontal distance. There is uncertainty about how to correctly apply these relationships, particularly regarding the role of the triangle formed by the motion.

Discussion Status

Some participants have suggested methods for calculating the height using tangent functions, while others have raised concerns about the assumptions made regarding the triangle's dimensions. There is an ongoing exploration of the relationship between the vertical and horizontal components of the velocity at impact.

Contextual Notes

Participants are navigating potential misunderstandings about the geometry of the problem and the definitions of the components involved. The discussion reflects a mix of interpretations regarding the setup and the application of kinematic equations.

sabres
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A ball is thrown leftward from the left edge of the roof of a building at height h above the ground. The ball hits the ground 1.50s later, at distance d=25.0m from the building and at angle theta = 60 degrees with the horizontal.

A) Find h.

What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown?




I figured that initial velocity in X direction is 16.67 m/s (25 meters/1.5 seconds). What is really giving me trouble is finding the magnitude of the velocity. I can't figure out how to find the Y component of the velocity to get it. I'm not even sure what B and C want, but I figure i'll cross that bridge after I get A handled.
 
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Use the angle given 60 degrees, to find the height.
 
Could you be more specific please? I don't know how to do that.
 
If I am thinking of this right then your angle is opposite of your horizontal so you use tan(60)x25.0 m to find the horizontal.
 
Fizicks said:
If I am thinking of this right then your angle is opposite of your horizontal so you use tan(60)x25.0 m to find the horizontal.

yes, tan(60)*25 is the height h.
 
That is what I thought originally, too. Problem is the 25 is not the hypotenuse of the triangle that is formed, so (tan 60) * 25 yields an incorrect answer. That gives 43.3, but the back of the book says it is 32.3
 
sabres said:
That is what I thought originally, too. Problem is the 25 is not the hypotenuse of the triangle that is formed, so (tan 60) * 25 yields an incorrect answer. That gives 43.3, but the back of the book says it is 32.3

Ah... looks like I misunderstood the problem. The object is thrown at an angle, so it has an initial vertical velocity... it hits the ground with an angle of 60... that means that (vertical speed)/(horizontal speed) is tan(60) when the object hits.

ie it hits with a vertical velocity of - 16.67tan(60) = -28.87m/s

using this you can get the height, and it comes out to 32.3m.
 
You're a lifesaver, thank you!
 

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