# Help me with the question?

Sep 27, 2017

$\alpha = {60}^{o}$

#### Explanation:

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

$\left.\begin{matrix}v = u + a t & \text{ where " & s="displacement "(m) \\ s=ut+1/2at^2 & \null & u="initial speed "(ms^-1) \\ s=1/2(u+v)t & \null & v="final speed "(ms^-1) \\ v^2=u^2+2as & \null & a="acceleration "(ms^-2) \\ s=vt-1/2at^2 & \null & t="time } \left(s\right)\end{matrix}\right.$

Let us define the following variables:

 { (U, "= Initial Speed", ms^-1), (alpha, "= Angle of projection to the horizontal", ""^o), (h, "= Maximum Height", m), (U_h, "= Horizontal Speed (constant)", ms^-1), (V_("top"), "= Overall speed at highest point", ms^-1), (V_("half"), "= Overall speed at half of maximum height", ms^-1) :}

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with $a = 0$). This horizontal speed is given by

${U}_{h} = U \cos \alpha$

Vertical Motion

The projectile travels under constant acceleration due to gravity.

When the projectile reaches its maximum height Its vertical speed will momentarily be zero, so we can calculate the maximum height:

$\left\{\begin{matrix}s = & h & m \\ u = & U \sin \alpha & m {s}^{-} 1 \\ v = & 0 & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & \text{Not Required} & s\end{matrix}\right.$

Applying ${v}^{2} = {u}^{2} + 2 a s$ we have:

${0}^{2} = {\left(U \sin \alpha\right)}^{2} + \left(2\right) \left(- g\right) \left(h\right)$
$\therefore 2 g h = {U}^{2} {\sin}^{2} \alpha$
$\therefore h = \frac{{U}^{2} {\sin}^{2} \alpha}{2 g}$

The overall speed at this maximum height is just the horizontal component, thus:

${V}_{\text{top}} = U \cos \alpha$

Now, let us calculate the vertical speed at halfway of the maximum height, ie $\frac{h}{2}$

$\left\{\begin{matrix}s = & \frac{{U}^{2} {\sin}^{2} \alpha}{4 g} & m \\ u = & U \sin \alpha & m {s}^{-} 1 \\ v = & v & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & \text{Not Required} & s\end{matrix}\right.$

Applying ${v}^{2} = {u}^{2} + 2 a s$ we have:

${v}^{2} = {\left(U \sin \alpha\right)}^{2} + 2 \left(- g\right) \left(\frac{{U}^{2} {\sin}^{2} \alpha}{4 g}\right)$
$\setminus \setminus \setminus = {U}^{2} {\sin}^{2} \alpha - \frac{{U}^{2} {\sin}^{2} \alpha}{2}$
$\setminus \setminus \setminus = \frac{{U}^{2} {\sin}^{2} \alpha}{2}$

So, the overall speed at this half way height is given by:

${V}_{\text{half}}^{2} = {v}^{2} + {U}_{h}^{2}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{U}^{2} {\sin}^{2} \alpha}{2} + {U}^{2} {\cos}^{2} \alpha$

$\therefore {V}_{\text{half}} = \sqrt{\frac{{U}^{2} {\sin}^{2} \alpha}{2} + {U}^{2} {\cos}^{2} \alpha}$

The next step is to use the fact that the speed at the maximum height is $\sqrt{\frac{2}{5}}$ the speed at half height:

$\implies {V}_{\text{top") = sqrt(2/5) xx V_("half}}$
$\therefore {V}_{\text{top")^2 = 2/5 xx V_("half}}^{2}$

And using the values determined by the above calculation:

${\left(U \cos \alpha\right)}^{2} = \frac{2}{5} \left\{\frac{{U}^{2} {\sin}^{2} \alpha}{2} + {U}^{2} {\cos}^{2} \alpha\right\}$

$\therefore \frac{5 {U}^{2} {\cos}^{2} \alpha}{2} = \frac{{U}^{2} {\sin}^{2} \alpha}{2} + {U}^{2} {\cos}^{2} \alpha$

$\therefore \frac{5 {\cos}^{2} \alpha}{2} = \frac{{\sin}^{2} \alpha}{2} + {\cos}^{2} \alpha$

$\therefore 5 {\cos}^{2} \alpha = {\sin}^{2} \alpha + 2 {\cos}^{2} \alpha$

$\therefore 3 {\cos}^{2} \alpha = {\sin}^{2} \alpha$

$\therefore {\sin}^{2} \frac{\alpha}{\cos} ^ 2 \alpha = 3$

$\therefore {\tan}^{2} \alpha = 3$

$\therefore \tan \alpha = \pm \sqrt{3}$

However, as $\alpha$ must be acute we have:

$\tan \alpha = \sqrt{3}$

Which then gives us:

$\alpha = {60}^{o}$