Projectile motion of a baseball problem

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Homework Help Overview

The problem involves the projectile motion of a baseball thrown at an angle of 25 degrees with an initial speed of 23.0 m/s. The original poster seeks to determine the time the ball was in the air and the maximum height reached before being caught 42 m away.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use kinematic equations to find the time of flight and maximum height but questions their calculations after receiving feedback indicating an error. Some participants suggest reviewing the calculations for the vertical motion and the interpretation of the results.

Discussion Status

Participants are actively discussing the original poster's approach and calculations. Some guidance has been offered regarding the interpretation of the equations used, particularly concerning the vertical motion and the significance of the maximum height.

Contextual Notes

There is mention of a discrepancy between the original poster's calculations and the expected results from a reference source, prompting a review of assumptions and methods used in the calculations.

princesspriya
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Homework Statement


a baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23.0m/s. If the ball was caught 42 m from the thrower, how long was it in the air? how high did the ball travel before being caught?

Homework Equations




The Attempt at a Solution



using x=(VicoxA)t --> 42/(23*cos25) i got 4.32
and than using y=(23*sin25)4.32+1/2(-9.81)(4.32)^2 i got -49.5 >>y=(VisinA)t+1/2a(t)^2)
but in the back of the book it says its wrong so can someone please tell me what i did wrong?
 
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using x=(VicoxA)t --> 42/(23*cos25) i got 4.32
and than using y=(23*sin25)4.32+1/2(-9.81)(4.32)^2 i got -49.5 >>y=(VisinA)t+1/2a(t)^2)
but in the back of the book it says its wrong so can someone please tell me what i did wrong?[/QUOTE

The formula for projectile motion is the Y direction is going to be delta y=Vt -1/2gt^2 assuming gravity is -9.8. As a result of this double negative you end up adding the latter half of the equation where it appears you have subracted by adding a negative. Give that a try and see if it gives you a better answer
 
well...look at your second step: you are not calculating the hightest point...(and I'm not sure but -49.5 is not right: you are saiyng that y=-49.5?? the ball must be underground...xD)

hint: In the hightest point, the ball as no speed in y axis...^_^
 
ooo thanks a lot!
 

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