How do i find the range?

  • #1

Homework Statement


a soccer player kicks a stationary ball at a speed of 20m/s at an angle of 15[tex]\circ[/tex] 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 plz!!
 

Answers and Replies

  • #2
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?
 
  • #3
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)????
 
  • #4
Whats the acceleration in the horizontal direction?

hint. if you ignore air resistance what forces act horizontally
 
  • #5
No idea...im sooo lost!!!!! i looked in my book and found this formula R=Vo(square)sin2teta/g omg im freaking out! what Vo i have to take?!? 20m/s? the Vy? or the Vx?
 
  • #6
is it (19.32m/s)^2 sin(30)/9.8m/s^2= 19.04m?
 
  • #7
hmm...the only force i see is the Vx=20m/s cos 15=19.32m/s it's the only one horizontaly
 
  • #8
754
1
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).
 
  • #9
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
 

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