Angle, horizontal range, velocity

[marinabradaric]

Homework Statement

A soccer player kicked a ball up at a 30-degree angle at 20 m/s. Find the hang time, max. height and horizontal range.

Homework Equations

x(sub y)=(1/2)at^2 + v(sub i sub y)
Meaning... displacement = half of acceleration times time squared plus initial velocity in the y-direction.
v(sub f)^2 = v(sub i)^2 + 2ax
delta X = v(sub x) * T (sub y)

The Attempt at a Solution

For the first equation, x(sub y)=(1/2)at^2 + v(sub i sub y)
displacement is 0 because horizontally, it goes from the ground and ends back at the ground; no change. half of acceleration of -9.8 m/2^2 is 4.9 m/s^2. Now for the initial velocity, it says that it's in the y direction.. so you're supposed to set up a triangle from the curved path of the ball with the angle of 30 degrees, x on the bottom, y on the opposite of the 30d, with the hypotenuse being 20 m/s. When you take the sin 30 = y/20, you get y=10, so velocity in the y direction is ten. First of all, this was unclear to me because if you draw the triangle, it looks like y should be the maximum height. Anyway, so you plug those numbers into the equation, giving you.
0=-4.9(m/s^2) * t^2 + 10 (m/s)
-10=-4.9t^2
2.04082 = t^2
1.43=t

But the answer is supposed to be two seconds. Am I missing something here?

Then after that, you want to find the max. height so you take
v(sub f)^2 = v(sub i)^2 + 2ax and plug in numbers.
0 = 10^2(m/s) + 2(-9.8 m/s^2) * x
0 = 100 m/s - 19.6x
-100 m/s = -19.6x
5.10 m = x

Now comes the horizontal range, when you use delta X = v(sub x) * T(sub y)
To find velocity in the x direction, you go to the triangle mentioned above and use cosine of 30 degrees instead, ending up with x=17.3 m/s.
So delta x = 17.3 m/s * 2 s, which gives you delta x=34.6 m but everyone else got 35.35. Again, am I missing something?

Thanks for the help!

 

[tony873004]

You are correct that the y-component of the velocity equals 10 m/s. sin(30)=0.5. 0.5*20=10.

Hint: instead of jumping straight to the formulas, just try to intuitively picture what is happening:

You know that gravity slows things down by 10 meters per second every second.

How long will it take something that is traveling 10 meters per second to be slowed to a stop if it is decelerating at 10 meters per second every second?

That gets you to the top. Double it to get the ball back to the ground. You need this answer before you can find the other parts.

 

[m2003]

I would like to clarify a few points in your attempt at solving the problem. Firstly, the equation x(sub y)=(1/2)at^2 + v(sub i sub y) is only applicable for objects in free fall, where there is a constant acceleration of -9.8 m/s^2. In this case, the ball is not in free fall as it is kicked at an initial velocity of 20 m/s. Therefore, this equation cannot be used to find the time of flight.

To find the time of flight, we can use the equation t = v(sub f)/g, where v(sub f) is the final velocity in the y direction and g is the acceleration due to gravity, which is -9.8 m/s^2. In this case, v(sub f) can be found by using the equation v(sub f) = v(sub i) + at, where v(sub i) is the initial velocity in the y direction and a is the acceleration due to gravity. Plugging in the values, we get v(sub f) = 10 m/s. Therefore, t = 10/9.8 = 1.02 seconds, which can be approximated to 1 second.

Moving on to finding the maximum height, we can use the equation h = v(sub i)^2/2g, where h is the maximum height reached. Again, v(sub i) is the initial velocity in the y direction and g is the acceleration due to gravity. Plugging in the values, we get h = (10)^2/(2*9.8) = 5.10 meters, which is the correct answer.

For the horizontal range, we can use the equation R = v(sub i)*t, where R is the horizontal range and t is the time of flight. In this case, since we have already found the time of flight to be 1 second, we can use that value to find the horizontal range. Plugging in the value of v(sub i) = 17.3 m/s (which is the horizontal component of the initial velocity of 20 m/s at an angle of 30 degrees), we get R = 17.3*1 = 17.3 meters, which can be approximated to 17.3 meters.

I hope this clarifies your doubts and helps you understand the problem better. As scientists, it is important for us to