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

[physkid1]

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 ??

 

[Hootenanny]

We can't help you if we don't know the question!

 

[physkid1]

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

 

[Hootenanny]

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?

 

[physkid1]

i understand a quadratic has 2 solutions but i can't seem to see how that equation is a quadratic

 

[Hootenanny]

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!

 

[iRaid]

Anything thrown upwards (at least on the earth) will have a parabolic shape - which is a quadratic.

 

[PeterO]

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]

 

[c03rcion]

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, t_A and t_B.
t_A is when the object is at the peak of the parabola.
t_B is when the object lands.

t_B = 2t_A

**since acceleration caused by gravity is constant
A_y = (V_f - V_i) / t_A

A_y = -g

-gt_A + V_i = V_f

**V_f=0 because A_y is ZERO at the apex of the parabola

t_A = -V_i/-g =1.42s

THE TOTAL TIME IN FLIGHT IS t_B = 2t_A
2t_A = 2.9s = t_B

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.