Calculating Height of Ball Thrown at 30 Degrees

AI Thread Summary
A ball thrown at a 30-degree angle with an initial speed of 6 m/s takes 3 seconds to hit the ground, prompting a calculation for its launch height. The initial attempt used the equation yf = yi + Viy * t - 0.5 * g * t^2, resulting in a negative height, indicating the ball lands below the launch point. Clarification revealed that the correct interpretation involves considering both vertical and horizontal components of motion. The discussion highlighted the need for proper equations to find maximum height and final velocity accurately. Overall, the importance of correctly applying kinematic equations in projectile motion was emphasized.
-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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