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.:Endeavour:.
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Homework Statement
I'm still kind of confused with projectile motion. I know when you throw something, you throw it at an angle and will have a negative parabola effect. An exception to when you throw something up at a 90° angle where it will go up vertically and come down vertically, with no air resistance.
The problem is:
A ball is launched at a 4.5 m/s at 66° above the horizontal. What are the maximum height and flight time of the ball?
Homework Equations
ymax = yi + vyit + [tex]\frac{1}{2}[/tex]*at2
possibly:
vf = vi2 + 2*a(df - di)
The Attempt at a Solution
This is an example from the book which doesn't clearly explains it. The first step is to find the x and y component.
My variables are:
yi = 0 meter
vi = 4.5 m/s
[tex]\theta[/tex]i = 66°
acceleration (a) = -g
I want to find:
ymax = ?
t = ?
Because the projectile is traveling in a parabola form, the way to find the total time traveled by the projectile is by using the quadratic formula which is:
[tex]
x = \frac{-b +/- \sqrt{b^2 - 4ab}}{2a}
[/tex]
where
+/- = ±
c = yi
b = vyi
a = is Fg
x = time
I know where they got the vyi by using the formula:
vyi = vi(sin [tex]\theta_i[/tex])
vyi = (4.5m/s)(sin 66°)
vyi = 4.1 m/s
This is the formula that they used to solve for the height:
ymax = yi + vyit + [tex]\frac{1}{2}[/tex]*at2
I'll like some more clarification on that formula and the quadratic formula that the book used to find the time. If I wanted to find the distance traveled by the projectile how do I find it? Do I use this formula to find the distance traveled by the projectile:
vf = vi2 + 2*a(df - di)
Thank you for your time.