
#1
Sep2612, 05:30 PM

P: 9

1. A soccer ball is kicked so that it has a range of 30m and reaches a maximum height of 12m. What velocity (magnitude and direction) did the ball have as it left the footballer's foot?
I used the equation to solve for time and got 1.55s. I'm not sure if air resistance is involved, we never discussed how to solve problems with air resistance in class. 2. A stone is thrown with a speed of 20.0 m/s at an angle of 48° to the horizontal from the edge of a cliff 60.0m above the surface of the sea. (a) Calculate the velocity with which the stone hits the sea. (b) Discuss qualitatively the effect of air resistance on your answers to (a). 3. The maximum height reached by a projectile is 20m. The direction of the velocity 1.0s after launch is 20°; find the speed of launch. Is air resistance included in all of these problems? 



#2
Sep2612, 05:46 PM

PF Gold
P: 1,054

For the first problem, what does 1.55s represent? Air resistance is assumed negligible.




#3
Sep2612, 06:08 PM

P: 9

Sorry, s stands for seconds. Could you explain how to solve the problems?




#4
Sep2612, 06:18 PM

PF Gold
P: 1,054

Projectile Motion Problems
Sorry, I meant the interval of time, in seconds, what does that represent? Which equation did you use? Show those steps. For example of what I'm asking: does your time result represent the total flight time or the time it took to reach the highest point during flight?




#5
Sep2612, 06:48 PM

P: 9

I think the time is how long it was in flight, I'm not sure if this is correct or if I was even supposed to solve for time. I used this equation:
Δy=Vyt + 0.5αyt^2 12=0.5(10)(t^2) and used this to solve for t 



#6
Sep2612, 06:55 PM

PF Gold
P: 1,054

Okay, there was a term, Vyt. How did you make that disappear?




#7
Sep2612, 06:58 PM

P: 9

Isn't the velocity of y zero? So then it would be (0)(1.55) which is zero, right?




#8
Sep2612, 07:11 PM

PF Gold
P: 1,054

So to visualize what is going on along the y axis initially the footballer's foot gives an initial V_{y0} resulting in the ball going up. At some time, the ball stops going up V_{y} = 0. This is what made V_{y}t disappear in your equation. You have solved for this time. Now the ball comes back down. The moment before it hits the ground, it has a nonzero velocity. Do not consider this zero velocity. By symmetry you can determine the total flight time.




#9
Sep2612, 07:15 PM

P: 9

Ok then how do I use this to find the magnitude and direction of velocity?




#10
Sep2612, 07:23 PM

PF Gold
P: 1,054

Now that you know the total flight time, can you determine V_{x}?




#11
Sep2612, 07:41 PM

P: 9

I could use Vx=Vox+at but I don't know what Vox is.




#12
Sep2612, 07:46 PM

PF Gold
P: 1,054

In the x direction, acceleration is 0. How about distance = rate * time?




#13
Sep2612, 07:55 PM

P: 9

I'm not sure what that formula is, will it help me solve for velocity? These are IB questions, not sure if that makes a difference.




#14
Sep2612, 08:00 PM

PF Gold
P: 1,054

That formula is a result of this formula with a = 0:
x  x_{0} = v_{0}t+ 0.5at^{2} 



#15
Sep2612, 08:05 PM

P: 961

I think should start with what you know and do not know that lead to the answer.
Knowns(3)  x and y(0 and 12m) distances. Unknowns(3)  v_{0}, θ, t So you need 3 equations. There are 3 SUVAT equations. s=v_{0}t + 1/2 at^{2} v^{2}=u^{2}+2as vu =at In this problem we are taking up positive and acceleration a as negative. You have 2 s(x and y) which are independent. So you need 3rd equation. You can use the last equation(v_{y}u_{y})=at as the third equation(only in y direction with acceleration). 



#16
Sep2612, 08:15 PM

PF Gold
P: 1,054




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