Optimizing Thrown Ball Trajectories

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Homework Help Overview

The discussion revolves around the physics of projectile motion, specifically focusing on the trajectory of a thrown ball in different gravitational environments. The original poster presents a scenario involving an astronaut throwing a ball on Earth and seeks to determine the maximum height achieved when thrown at an optimal angle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum height using initial velocity and angle but expresses confusion regarding the results. Some participants suggest revisiting the initial velocity used in the calculations.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to the problem. There is acknowledgment of a potential error in the initial velocity used for calculations, but no consensus has been reached on the correct method or resolution of the height calculation issue.

Contextual Notes

The original poster is working under the assumption that the angle for maximum range is 45 degrees and is also considering the effects of a different gravitational acceleration on another planet. There is a noted difficulty in applying the equations of motion correctly, particularly regarding vertical acceleration.

Bryon
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Homework Statement



An astronaut in his space suit can throw a ball a maximum distance dmax = 9 m on the surface of the earth.

For a given speed of the ball, what angle to the horizontal q (in degrees) will yield the greatest range? 45 degrees

If the ball is thrown at this same angle q, what speed will produce this greatest range (9 m) ? 9.3m/s

How far can he throw the ball on a planet where g1 = 22 m/s2? 4.009

What height will the ball reach on this "maximum range" trajectory? (on the planet where g1 = 22 m/s2)? I am having a problem with this one. I need help!


Homework Equations



v(y) = v(initial)*sin(angle)
v(y) = v( y initial) + at
y = y(initial) + 0.5(v(inital) + v)t


The Attempt at a Solution



v(y) = 9sin(45) = 6.57

0 = 6.57 + 22(t)
t = 0.298

y = 0 + 0.5(6.57 + 0)*(0.298)
y = 0.978

Where did I go wrong? Did I need the vertial acceleration?

thanks!
 
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(y) = 9sin(45) = 6.57
It should be
(y) = 9.3sin(45)
 
Ah yes I made a type, it should be 9.3. But I still am having trouble finding the maximum height...It seems that every approach I've tried it comes up wrong.
 
Bryon said:
Ah yes I made a type, it should be 9.3. But I still am having trouble finding the maximum height...It seems that every approach I've tried it comes up wrong.
t = 0.2986
= 0.299
 

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