# Projectile motion

1. Oct 1, 2008

### -EquinoX-

1. The problem statement, all variables and given/known data
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 [Broken]http://g.imageshack.us/thpix.php [Broken]

2. Relevant equations

3. 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: May 3, 2017
2. Oct 1, 2008

### LowlyPion

The number is negative because it lands below the point it was thrown.

Last edited by a moderator: May 3, 2017
3. Oct 1, 2008

### -EquinoX-

so the answer is right 37 ? or is it supposed to be 9+45?

4. Oct 1, 2008

### LowlyPion

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.

5. Oct 1, 2008

### -EquinoX-

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) ??

6. Oct 1, 2008

### LowlyPion

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.

7. Oct 1, 2008

### -EquinoX-

so what equation should I use here?

8. Oct 1, 2008

### LowlyPion

Was that asked as part of the question?

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