# Forming a quadratic equation/Physics motion quesiton?

• Jun 10th 2008, 12:45 PM
thermalwarrior
Forming a quadratic equation/Physics motion quesiton?
A particle is observed to to pass through a distance d horizontally and d/10 vertically from the point of projection. Show that if T= tan (theta) then T satisfies the equation:

75T^2 - 40T + 4 =0

What is confusing me is how to do this without values as I have an equation to form the quadratic equation:

x tan(theta) - x^2(g/2u^2)sec^2(theta)
• Jun 10th 2008, 01:29 PM
topsquark
Quote:

Originally Posted by thermalwarrior
A particle is observed to to pass through a distance d horizontally and d/10 vertically from the point of projection. Show that if T= tan (theta) then T satisfies the equation:

75T^2 - 40T + 4 =0

What is confusing me is how to do this without values as I have an equation to form the quadratic equation:

x tan(theta) - x^2(g/2u^2)sec^2(theta)

Here's a partial solution anyway. Assume that $\theta$ is the projection angle. Then if you take the ratio of the y component of the initial velocity and the x component of the initial velocity you get $tan(\theta)$. So write out the x equation of motion and the y equation of motion:
$d = v_0~cos(\theta)~t$

$\frac{d}{10} = v_0~sin(\theta)~t - \frac{1}{2}gt^2$

Thus
$tan(\theta) = \frac{v_0~sin(\theta)}{v_0~cos(\theta)} = \frac{\frac{d}{10t} + \frac{1}{2}gt}{\frac{d}{t}}$

$tan(\theta) = \frac{1}{10} + \frac{g}{2d}t^2$

From the cosine equation
$t = \frac{d}{v_0~cos(\theta)}$

so
$tan(\theta) = \frac{1}{10} + \frac{g}{2d} \left ( \frac{d}{v_0~cos(\theta)} \right )^2$

$tan(\theta) = \frac{1}{10} + \frac{gd}{2v_0^2}~sec^2(\theta)$

$tan(\theta) = \frac{1}{10} + \frac{gd}{2v_0^2}~(1 + tan^2(\theta))$

At the moment I can't think of a way to eliminate the $v_0$ from this.

-Dan
• Jun 10th 2008, 01:42 PM
thermalwarrior
i think it maybe easier than it sounds as the question is only 6 marks and this includes find the values of the quadratic equations.

This was the previous part of the question:

An object of mass M is projected from the floor of a horizontal tunnel of height 2d/15 with an initial speed of u at an angle (theta) above the horizontal.

the object just touches the tunnel roof show that:

u^2 = 4dg/15sin^2(theta)

Now ive done that. So can i use this equation to form the quadratic equation???
• Jun 10th 2008, 04:12 PM
topsquark
Quote:

Originally Posted by thermalwarrior
i think it maybe easier than it sounds as the question is only 6 marks and this includes find the values of the quadratic equations.

This was the previous part of the question:

An object of mass M is projected from the floor of a horizontal tunnel of height 2d/15 with an initial speed of u at an angle (theta) above the horizontal.

the object just touches the tunnel roof show that:

u^2 = 4dg/15sin^2(theta)

Now ive done that. So can i use this equation to form the quadratic equation???

Ah, good. That gives us what we need. Your u is my $v_0$, so continuing:
$tan(\theta) = \frac{1}{10} + \frac{gd}{2v_0^2}sec^2(\theta)$

with
$v_0^2 = \frac{4dg}{15~sin^2(\theta)}$

gives
$tan(\theta) = \frac{1}{10} + \frac{gd}{2 \left ( \frac{4dg}{15~sin^2(\theta)} \right ) }sec^2(\theta)$

$tan(\theta) = \frac{1}{10} + \frac{15}{8}sin^2(\theta)~sec^2(\theta)$

$tan(\theta) = \frac{1}{10} + \frac{15}{8}tan^2(\theta)$

$40~tan(\theta) = 4 + 75~tan^2(\theta)$

or finally:
$75~tan^2(\theta) - 40~tan(\theta) + 4 = 0$

-Dan
• Jun 17th 2008, 10:29 AM
nikhil
Very simple
Hi, thermalwarrior.
This is a very easy question.
Horizontal range is given by the formula=>[u^2sin(2@)]/g
and vertical height by the formula =>(usin(@))^2/2g.
[@=theta and u=initial velocity]
According to the question
d=[u^2sin(2@)]/g and
d/10=(usin(@))^2/2g
dividing both we get
tan@=2/5
[sin2@=2sin@cos@]
putting value of tan@ in the given expression we get
75(2/5)^2-40(2/5)+4=0 hence proved.