Calculating Height of Ball Thrown at 30 Degrees

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Homework Help Overview

The discussion revolves around calculating the height from which a ball is thrown at an angle of 30 degrees with an initial speed of 6 m/s, given that it takes 3 seconds to hit the ground. Participants are exploring the implications of their calculations and the physics involved in projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to use the kinematic equation for vertical motion to find the height, questioning the negative result obtained. They explore the implications of the ball landing below the initial height and discuss the correctness of their calculations.

Discussion Status

The discussion is ongoing, with participants questioning the validity of their approaches and calculations. There is a mix of interpretations regarding the results, and some guidance has been offered about the need to consider both components of velocity.

Contextual Notes

Participants are grappling with the constraints of the problem, including the initial conditions and the equations applicable to projectile motion. There is uncertainty regarding the correct interpretation of the results and the equations to use for further calculations.

-EquinoX-
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Homework Statement


A ball is thrown at the angle 30 degrees with an initial speed of 6 m/s, it takes the ball 3 seconds to hit the ground. Calculate the height where the ball was thrown
http://img355.imageshack.us/img355/2407/phyuq7.th.jpg http://g.imageshack.us/thpix.php

Homework Equations





The Attempt at a Solution



I tried to solve this using the equation:

yf = yi + Viy *t - 0.5 * g * t^2
yf-yi = (6 * sin(30) * 3 s) - (0.5*10*3^2)
= 9 - 45
= -37

why is it negative? What am I doing wrong here?
 
Last edited by a moderator:
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-EquinoX- said:

Homework Statement


A ball is thrown at the angle 30 degrees with an initial speed of 6 m/s, it takes the ball 3 seconds to hit the ground. Calculate the height where the ball was thrown
http://img355.imageshack.us/img355/2407/phyuq7.th.jpg http://g.imageshack.us/thpix.php

Homework Equations





The Attempt at a Solution



I tried to solve this using the equation:

yf = yi + Viy *t - 0.5 * g * t^2
yf-yi = (6 * sin(30) * 3 s) - (0.5*10*3^2)
= 9 - 45
= -37

why is it negative? What am I doing wrong here?

The number is negative because it lands below the point it was thrown.
 
Last edited by a moderator:
so the answer is right 37 ? or is it supposed to be 9+45?
 
-EquinoX- said:
so the answer is right 37 ? or is it supposed to be 9+45?

It would be 9 + 45 if you threw it downward at the same speed and angle to the horizon.

By the way the correct answer for 45 - 9 is 36 and not 37.
 
ok, then to find the final velocity when the ball hits the ground I need to find:

Vyf^2 = Vyi^2 -2gh
= (6* sin 30)^2 - 2*(10)*(45)
= 9 - 900
= sqrt (-891) ??
 
-EquinoX- said:
ok, then to find the final velocity when the ball hits the ground I need to find:

Vyf^2 = Vyi^2 -2gh
= (6* sin 30)^2 - 2*(10)*(45)
= 9 - 900
= sqrt (-891) ??

That would be because that is not a correct equation. For instance are you saying that if initial velocity was 30 m/s upwards, that by your equation that final velocity would be 0?

To use that equation you would need to find the maximum height first, because what you are using describes a velocity translation in one direction.
 
so what equation should I use here?
 
-EquinoX- said:
so what equation should I use here?

Was that asked as part of the question?

Because if it was you have to use both components of velocity.
 

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