How do i find the range?

[yellow_angel]

Homework Statement

a soccer player kicks a stationary ball at a speed of 20m/s at an angle of 15$$\circ$$ to the horizontal.

Homework Equations

i think it would be Rmax=Vo²/g ?

The Attempt at a Solution

19.32 ²/9.8=38.09m ??

I'm really lost can someone help me please!

 

[NobodySpecial]

From the angle work out the vertical speed
Then think about how the vertical speed changes with time

hint. what happens at the top of the flight?

 

[yellow_angel]

mmm... so i calculate the vertical speed: 20sin15=5.18m/s? and with that i can find at what time it reaches the max height by doing V=Vo+at: V=5,18/9.8= 0.53s do i have to multiply the time by 2 and then do the same thing to find the horizontal range?

X=Xo+Vot-1/2gt(square) where g=0 and Xo=0 too, so it would be X=19.32*(2*0.53s)?

 

[NobodySpecial]

Whats the acceleration in the horizontal direction?

hint. if you ignore air resistance what forces act horizontally

 

[yellow_angel]

No idea...im sooo lost! i looked in my book and found this formula R=Vo(square)sin2teta/g omg I am freaking out! what Vo i have to take?!? 20m/s? the Vy? or the Vx?

 

[yellow_angel]

is it (19.32m/s)^2 sin(30)/9.8m/s^2= 19.04m?

 

[yellow_angel]

hmm...the only force i see is the Vx=20m/s cos 15=19.32m/s it's the only one horizontaly

 

[zgozvrm]

mmm... so i calculate the vertical speed: 20sin15=5.18m/s? and with that i can find at what time it reaches the max height by doing V=Vo+at: V=5,18/9.8= 0.53s do i have to multiply the time by 2 and then do the same thing to find the horizontal range?

X=Xo+Vot-1/2gt(square) where g=0 and Xo=0 too, so it would be X=19.32*(2*0.53s)?

All of this is correct!
(Although, you didn't show where or how you got 19.32).

 

[NobodySpecial]

hmm...the only force i see is the Vx=20m/s cos 15=19.32m/s it's the only one horizontaly

Correct, there is no horizontal force, so no aceleration, so speed is constant