# Height and speed of a launched ball after two seconds [closed]

I have this problem, I have tried to do it but, without success, the given elements are very few:

A player launches a ball with an angle of $\alpha = 60°$ above the horizontal. After $t^{\ast} = 2 \, s$ (seconds), the ball still rises and its velocity has an angle of $\theta = 30°$ above the horizontal. Neglecting the air resistance, find:

a. the speed at the launching;
b. the height of the ball, with respect to the point of launching, after $t^{\ast} = 2 \, s$.

[answers : $\quad |v_0| = 34 \frac{m}{s}, \quad h = 39.3 \, m$]

I post what I have done, I have any other idea on how to solve it:

this is the image I have done:

I find the components, but the module still remains unknown:

$\begin{array}{lcl} v_{0_x} & = & |\vec{v_0}| \cdot \cos \alpha & = & |\vec v_0| \cdot \cos 60° & = & |\vec{v_0}| \cdot \frac{1}{2} \\ v_{0_y} & = & |\vec{v_0}| \cdot \sin \alpha & = & |\vec v_0| \cdot \sin 60° & = & |\vec{v_0}| \cdot \frac{\sqrt{3}}{2} \end{array}$

considering the following system of equations respectively for uniform rectilinear motion and for the uniformly accelerated motion:

$\left \{ \begin{array}{lcl} v_{0_x} & = & \frac{\Delta x}{\Delta t} \\ \Delta h & = & v_{0_y} \cdot \Delta t - \frac{1}{2} \cdot g \cdot (\Delta t)^2 \end{array} \right.$

where $\Delta x$ could be the space travelled in $\Delta t$, i.e. the time spent to travel that space till $2 \, s$. So supposing that starts from istant $0$, $\Delta t = t^{ \ast } - t_0 = (2 - 0)\, s = 2 \, s$.

$\Delta h$ is the height after two seconds.

if I go to substitute in the second equation :
$\begin{array}{lcl} h & = & \frac{\sqrt{3}}{3} |\vec{v_0}| \cdot 2 \, s - \frac{1}{2} \cdot 9.8 \frac{m}{s^2} \cdot 4 \, s^2 \\ & = & \sqrt{3} \cdot |\vec{v_0}| \cdot s - 19.62 \, m \end{array}$

but, from here I am unable to continue. Please, can you give me any suggestions? Thanks!

## closed as off-topic by Chris♦, Kyle Kanos, Sebastian Riese, John Rennie, stafusaMay 10 '18 at 8:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

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$tan(\theta) = \frac{usin(\alpha)-gt}{ucos(\alpha)}$ where $\alpha$ is angle of projection and $\theta$ is angle after time t. ( because at any instant $v_i = ucos(\alpha)$ and $v_j = usin(\alpha)-gt$ and angle is given by $tan(\theta)=\frac{v_j}{v_i}$ )
substituting the given values , we will get $u=20\sqrt3$ , and height after a given time is directly given by $h=usin(\alpha)t-\frac{gt^2}{2}$ which is 40m.
• sorry, I don't understand the $\tan \theta$. what is $u$, and what is $\phi$? Please, can you give me any additional clarification? Mant thanks! – JB-Franco May 11 '18 at 22:09
• $u$ is initial speed given, or $v_0$ w.r.t. your question. $\phi$ and $\theta$ are the angle of projection and angle made by the projectile after a time $t$ respectively . – 0xVikas May 12 '18 at 5:09
• does $\phi$ the one that I call in my exercise as $\alpha$? – JB-Franco May 12 '18 at 8:38