# Projectile Motion Problem?

Feb 15, 2017

a) $22.46$
b) $15.89$

#### Explanation:

Supposing the origin of coordinates at the player, the ball describes a parabola such as

$\left(x , y\right) = \left({v}_{x} t , {v}_{y} t - \frac{1}{2} g {t}^{2}\right)$

After $t = {t}_{0} = 3.6$ the ball hits the grass.

so ${v}_{x} {t}_{0} = {s}_{0} = 50 \to {v}_{x} = {s}_{0} / {t}_{0} = \frac{50}{3.6} = 13.89$

Also

${v}_{y} {t}_{0} - \frac{1}{2} g {t}_{0}^{2} = 0$ (after ${t}_{0}$ seconds, the ball hits the grass)

so ${v}_{y} = \frac{1}{2} g {t}_{0} = \frac{1}{2} 9.81 \times 3.6 = 17.66$

then ${v}^{2} = {v}_{x}^{2} + {v}_{y}^{2} = 504.71 \to v = 22.46$

Using the mechanical energy conservation relationship

$\frac{1}{2} m {v}_{y}^{2} = m g {y}_{\max} \to {y}_{\max} = \frac{1}{2} {v}_{y}^{2} / g = \frac{1}{2} {17.66}^{2} / 9.81 = 15.89$

Feb 15, 2017

$\textsf{\left(a\right)}$

sf(22.5color(white)(x)"m/s"

$\textsf{\left(b\right)}$

$\textsf{15.9 \textcolor{w h i t e}{x} m}$

#### Explanation:

$\textsf{\left(a\right)}$

Consider the horizontal component of the motion:

$\textsf{{V}_{x} = V \cos \theta = \frac{50.0}{3.6} = 13.88 \textcolor{w h i t e}{x} \text{m/s}}$

Since this is perpendicular to the force of gravity, this remains constant.

Consider the vertical component of the motion:

$\textsf{{V}_{y} = V \cos \left(90 - \theta\right) = V \sin \theta}$

This is the initial velocity of the ball in the y direction.

If we assume the motion to be symmetrical we can say that when the ball reaches its maximum height $\textsf{{t}_{\max} = \frac{3.6}{2} = 1.8 \textcolor{w h i t e}{x} s}$.

Now we can use:

$\textsf{v = u + a t}$

This becomes:

$\textsf{0 = V \sin \theta - 9.81 \times 1.8}$

$\therefore$$\textsf{V \sin \theta = 9.81 \times 1.8 = 17.66 \textcolor{w h i t e}{x} \text{m/s} = {V}_{y}}$

Now we know $\textsf{{V}_{x}}$ and $\textsf{{V}_{y}}$ we can use Pythagoras to get the resultant velocity V . This was the method used in the answer by @Cesereo R.

I did it using some Trig':

$\textsf{\frac{\cancel{v} \sin \theta}{\cancel{v} \cos \theta} = \tan \theta = \frac{17.66}{13.88} = 1.272}$

$\textsf{\theta = {\tan}^{- 1} 1.272 = {51.8}^{\circ}}$

This is the angle of launch.

Since $\textsf{{V}_{y} = V \sin \theta}$ we get:

$\textsf{V \sin \left(51.8\right) = 17.66}$

$\therefore$$\textsf{V = \frac{17.66}{\sin} \left(51.8\right) = \frac{17.66}{0.785} = 22.5 \textcolor{w h i t e}{x} \text{m/s}}$

$\textsf{\left(b\right)}$

To get the height reached we can use:

$\textsf{s = u t + \frac{1}{2} a {t}^{2}}$

This becomes:

$\textsf{s = V \sin \theta t - \frac{1}{2} \text{g} {t}^{2}}$

$\therefore$$\textsf{s = {V}_{y} t - \frac{1}{2} \text{g} {t}^{2}}$

Again, the time taken to reach the maximum height will be 3.6/2 = 1.8 s

$\textsf{s = 17.66 \times 1.8 - \frac{1}{2} \times 9.81 \times {1.8}^{2}}$ $\textsf{m}$

$\textsf{s = 31.788 - 15.89 = 15.9 \textcolor{w h i t e}{x} m}$