Solving Projectile Motion Equation from Physics 8e - Chapter 3 Page 70

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The discussion focuses on solving a projectile motion equation from "Physics 8e" by Cutnell and Johnson, specifically addressing the calculation of time of flight. The equation y = Voyt + 1/2 ayt^2 is highlighted, with the initial vertical velocity (Voy) at +14 m/s and acceleration due to gravity (ay) at -9.8 m/s^2. The time of flight is determined to be 2.9 seconds, derived from the understanding that the projectile's ascent and descent times are equal. Participants clarify that the equation is quadratic, which leads to two solutions, and emphasize that the total time in flight can be calculated using the peak time (tA) and doubling it for the descent (tB). The discussion concludes with a suggestion to refer to graphical representations of projectile motion for further clarity.
physkid1
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1. this is straight out of PHYSICS 8e by cutnell and johnson

chapter 3 page 70

there are two solutions to this equation. one is given by (14 m/s + 1/2 (-9.8 m/s ^2)t = 0 or t = 2.9s

what i can't seem to work out is how did they get t = 2.9s from those calculations




2. y = Voyt + 1/2 ayt^2



3. i just can't seem to calculate the final answer ... do i need to rearrange the equation more ??
 
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given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2
 
physkid1 said:
given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2
Correct so far. Since the equation you quote is quadratic, it has two solutions. The first is obviously t=0 (it starts off at ground level). This leaves

0 = v_0 + \frac{1}{2}at

or

t = -\frac{2 v_0}{a}

Do you follow?
 
i understand a quadratic has 2 solutions but i can't seem to see how that equation is a quadratic
 
physkid1 said:
i understand a quadratic has 2 solutions but i can't seem to see how that equation is a quadratic

y = v_0 t + \frac{1}{2}at^{\color{red}2}

The power "2" makes it a quadratic by definition!
 
Anything thrown upwards (at least on the earth) will have a parabolic shape - which is a quadratic.
 
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physkid1 said:
given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2

You could make use of the fact that the projectile come down just as fast as it went up, so final velocity is -14 m/s

That way you an use the non-quadratic formula v = u + at [not your symbols but a common set]
 
physkid1 said:
given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2

You don't need to use the quadratic equation.
I offer a more pictorial method for this problem.

There are two important times during the parabola trajectory, tA and tB.
tA is when the object is at the peak of the parabola.
tB is when the object lands.

tB = 2tA

**since acceleration caused by gravity is constant
Ay = (Vf - Vi) / tA

Ay = -g

-gtA + Vi = Vf

**Vf=0 because Ay is ZERO at the apex of the parabola

tA = -Vi/-g =1.42s

THE TOTAL TIME IN FLIGHT IS tB = 2tA
2tA = 2.9s = tB

Go to this link http://books.google.com/books?id=6u...resnum=1&ved=0CCEQ6AEwAA#v=onepage&q&f=false"

Go to page 80. You will see 2 graphs of parabola trajectory for projectile motion. Look at figure 4.9.
 
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