Projectile Motion parabola effect

[.:Endeavour:.]

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

y_max = y_i + v_yit + $$\frac{1}{2}$$*at²
possibly:
v_f = v_i² + 2*a(d_f - d_i)

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:
y_i = 0 meter
v_i = 4.5 m/s
$$\theta$$_i = 66°
acceleration (a) = -g

I want to find:
y_max = ?
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:
$$x = \frac{-b +/- \sqrt{b^2 - 4ab}}{2a}$$
where
+/- = ±
c = y_i
b = v_yi
a = is F_g
x = time

I know where they got the v_yi by using the formula:
v_yi = v_i(sin $$\theta_i$$)
v_yi = (4.5m/s)(sin 66°)
v_yi = 4.1 m/s

This is the formula that they used to solve for the height:
y_max = y_i + v_yit + $$\frac{1}{2}$$*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:
v_f = v_i² + 2*a(d_f - d_i)

Thank you for your time.

 

[LowlyPion]

First recognize that you can work the vertical and horizontal independently, since the vertical is accelerated uniformly by g, and the horizontal is not.

The equations you cite may or may not have all those terms depending on initial conditions and final position.

Sometimes you end up with quadratic expressions that are of course easily resolved with the quadratic equation.

For your problem then, if all you want to know is the max height and the total flight time, recognize some simplifying observations. The time to go up is the same as the time to go down and reach the same level it was thrown from.

The time to go up then is easily found by observing that since v = g*t, then you know that V_y initial divided by g is the time need until it reaches 0 velocity at the top. And twice that is the total time to go up and down.

Observe then that if you have the time for it to go up, that you can use that to find the max height directly with

y = 1/2*g*t²

 

[The Bob]

$$x = \frac{-b +/- \sqrt{b^2 - 4ab}}{2a}$$

Just a quick typo, 4ac not 4ab but I think you know that anyway.

The Bob

 

[.:Endeavour:.]

First recognize that you can work the vertical and horizontal independently, since the vertical is accelerated uniformly by g, and the horizontal is not.

The equations you cite may or may not have all those terms depending on initial conditions and final position.

Sometimes you end up with quadratic expressions that are of course easily resolved with the quadratic equation.

For your problem then, if all you want to know is the max height and the total flight time, recognize some simplifying observations. The time to go up is the same as the time to go down and reach the same level it was thrown from.

The time to go up then is easily found by observing that since v = g*t, then you know that V_y initial divided by g is the time need until it reaches 0 velocity at the top. And twice that is the total time to go up and down.

Observe then that if you have the time for it to go up, that you can use that to find the max height directly with

y = 1/2*g*t²

Ok I think I get what you are saying the only thing that keeps me puzzled is how the book explains it. This is how it explains it:

v_y = v_yi + a_yt
v_y = v_yi - gt

time = $$\frac{v_y_i - v_y}{g}$$

To sove for the maximum height, the book uses this formula:
y_max = y_i + v_yit + $$\frac{1}{2}$$ at²

y_max = $$y_i + v_y_i(\frac{v_y_i - v_y}{g}) + \frac{1}{2} (-g)(\frac{v_y_i - v_y}{g})^2$$

y_max =$$0.0 m + (4.1 m/s)(\frac{4.1 m/s - 0.0m/s}{9.80 m/s^2}) + \frac{1}{2}(-9.80 m/s^2)(\frac{4.1 m/s - 0.0 m/s}{9.80 m/s^2})^2$$

y_max = 0.86 meters

I'm still lost in this problem because where do you find v_y in the problem. For ballistics like this one problem, acceleration will be always negative when the projectile reaches its maximum height and then comes down at free-fall, negating wind resistance.

Yes, I think it was a typo. Its -4ac not -4ab. Thank you for your correction The Bob.

 

[LowlyPion]

Ok I think I get what you are saying the only thing that keeps me puzzled is how the book explains it. This is how it explains it:

v_y = v_yi + a_yt
v_y = v_yi - gt

These are the same equation. One is y positive down as when you drop a ball or roll down a ramp, the other is positive up as when you throw it up.

time = $$\frac{v_y_i - v_y}{g}$$

To sove for the maximum height, the book uses this formula:
y_max = y_i + v_yit + $$\frac{1}{2}$$ at²

You can use that formula and work it out or you can simplify it by understanding that the flight time up is the same as the flight time down when you throw something up from level ground. (If it is to end at a different height then you must use the whole equation.)

 

[.:Endeavour:.]

This is my final question. When you find the maximum height of the ball and also the time, can you plug in the time to the equation: y_max = $$y_i + v_y_i(\frac{v_y_i - v_y}{g}) + \frac{1}{2} (-g)(\frac{v_y_i - v_y}{g})^2$$ and still get the maximum height? The time is 0.84 seconds.

 

[LowlyPion]

This is my final question. When you find the maximum height of the ball and also the time, can you plug in the time to the equation: y_max = $$y_i + v_y_i(\frac{v_y_i - v_y}{g}) + \frac{1}{2} (-g)(\frac{v_y_i - v_y}{g})^2$$ and still get the maximum height? The time is 0.84 seconds.

That equation can also be expressed as

Viy² = 2*g*y

 

[.:Endeavour:.]

OK, thank you for your help, LowlyPion.