Difficult Projectile Question (2 Viewers)

Nicholasnewport · Mar 15, 2017

Question:

"a golfer hits a ball 135 metres on flat ground, 5m before it hits the ground it is at 1.8m high. Calculate the initial velocity"

Any help?

thanks for any advice in advance.

Nick

p.s. ignoring air resistance and all the usual stuff g=9.8m/s²

InteGrand · Mar 15, 2017

Nicholasnewport said:
Question:

"a golfer hits a ball 135 metres on flat ground, 5m before it hits the ground it is at 1.8m high. Calculate the initial velocity"

Any help?

thanks for any advice in advance.

Nick

p.s. ignoring air resistance and all the usual stuff g=9.8m/s²

$\noindent Let the initial speed be $V$ and angle of launch $\theta$ $\Big{(}$which is an angle between $0^{\circ}$ and $90^{\circ}$, i.e. between $0$ and $\frac{\pi}{2}$ in radians$\Big{)}$. Assume the launch occurs from the origin (where $x = 0$ and $y = 0$) as usual. By symmetry of parabolic flight, we equivalently have that the height is $y = 1.8$ at distance $5$ metres from the launching place (where $x = 5$).$

$\noindent The range $R$ of the projectile is given by $R = \frac{V^{2} \sin 2\theta}{g}$ (see a Wikipedia link below). We know that this is $135$, so we have$

$\boxed{\frac{V^{2} \sin 2\theta}{g} = 135. \quad (1)}$

$\noindent Now, we also know that $y = 1.8$ when $x = 5$. We know (see a Wikipedia link below) that the projectile's trajectory $y$ in terms of $x$ is given by$

$y = x \tan \theta - \frac{gx^{2}}{2V^{2}\cos^{2} \theta}.$

$\noindent So substituting $x = 5$ and $y = 1.8$ into this gives us$

$\boxed{5\tan \theta - \frac{25g}{2V^{2} \cos^{2} \theta} = 1.8. \quad (2)}$

$\noindent We thus have two equations in two unknowns ($V$ and $\theta$), which we can solve to find the initial velocity, so you can have a go at solving these now $\Big{(}$remember $V > 0$ and $0 < \theta < \frac{\pi}{2}$$\Big{)}$.$

$\noindent \textbf{Note.} For exam purposes, you will need to derive the equations I quoted above (shown in the Wikipedia links below), rather than just quote them. They are standard derivations that you should know how to do.$

$\noindent Links:$

- Range of projectile: https://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Distance_traveled
- Trajectory equation (with y₀ = 0): https://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Height_at_x .

mattskioner · Mar 15, 2017

As unhelpful as this is, the question appears to be outside the scope of the HSC physics course. Especially for prelim...

InteGrand · Mar 15, 2017

mattskioner said:
As unhelpful as this is, the question appears to be outside the scope of the HSC physics course. Especially for prelim...

It's from within the HSC maths syllabus (3U Projectile motion probably).

pikachu975 · Mar 15, 2017

InteGrand said:
It's from within the HSC maths syllabus (3U Projectile motion probably).

I think it shouldn't be in the prelim forum, projectile motion is HSC in both 3u and physics.

Difficult Projectile Question (2 Viewers)

Nicholasnewport

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InteGrand

Well-Known Member

mattskioner

New Member

InteGrand

Well-Known Member

pikachu975

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