Projectile Motion of Football: Solving for Time with Given Initial Conditions

[RedBurns]

A football is kicked at ground level with a speed of 40.0 m/s at an angle of 39.0° to the horizontal. How much later does it hit the ground?

X0= 0m Y0= 0m
X= ? Y= 0 m
Vx=? Vy= ?
Ax=0 m/s/s Ax=-9.8 m/s/s
T=?
V= 40.0 m/s
Theta= 39.0

Vx= 40.0 Cos (39.0) = 10.67 m/s
Vy= 40.0 Sin (39.0)= 38.55 m/s

0= 38.55T - 4.9 T^2

-38.55 +/- sqrt ((38.55)^2-4(-4.9)(0))/ 2(-4.9)
-38.55 +/- 38.55/-9.8
T= 0 and 7.87 seconds <-------------------------------I don't believe this is right. Could someone help me determine where I wondered astray?

 

[marlon]

Vx= 40.0 Cos (39.0) = 10.67 m/s

The 10.67 is incorrect.
40 * cos(39) = 31.1

Vy= 40.0 Sin (39.0)= 38.55 m/s

40*sin(39) = 25.2

0= 38.55T - 4.9 T^2

The "structure" and signs of this formula are correct though

marlon

 

[RedBurns]

Thanks! Just forgot to swithc the calculater from radians to degrees