Armstrong urgent!!!!!!

Hello Damien!

Neil Armstrong is on the moon (acceleration due to gravity is 1.6 m/s2) and throws a golf ball straight up with an upward with a speed of 10 m/s. How high does the golf ball go?

mv²/2 = m * a * h

h = v² / 2a

h = 10² m² / (s² * 2 * 1.6 m/s²)

h = 31.25 m

The golf ball rises to the height of 31.25 m.

Greeting asinus :- )

  !

Neil Armstrong is on the moon (acceleration due to gravity is 1.6 m/s2) and throws a golf ball straight up with an upward with a speed of 10 m/s. How high does the golf ball go?

$\begin{array}{|rcl|} \hline v_{\text{up}} &=& v_0 \\ v_{\text{down}} &=& g_{\text{moon}}\cdot t \\ \hline v &=& v_{\text{up}} - v_{\text{down}}\\ v &=& v_0 - g_{\text{moon}}\cdot t\\ \text{high max, if } v = 0 \\ 0 &=& v_0 - g_{\text{moon}}\cdot t\\ g_{\text{moon}}\cdot t &=& v_0 \\ \color{red}\mathbf{t} & \color{red}\mathbf{=} & \color{red}\mathbf{\frac{v_0}{g_{\text{moon}}} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline h_{\text{up}} &=& v_0\cdot t \\ h_{\text{down}} &=& \frac{g_{\text{moon}}} {2} \cdot t^2 \\ \hline h &=& h_{\text{up}} - h_{\text{down}}\\ h &=& v_0\cdot t - \frac{g_{\text{moon}}} {2} \cdot t^2 \qquad &| \qquad \color{red}\mathbf{t} & \color{red}\mathbf{=} & \color{red}\mathbf{\frac{v_0}{g_{\text{moon}}} }\\ h_{\text{max}} &=& v_0\cdot \left( \frac{v_0}{g_{\text{moon}}} \right) - \frac{g_{\text{moon}}} {2} \cdot \left(\frac{v_0}{g_{\text{moon}}} \right)^2 \\ h_{\text{max}} &=& \frac{v_0^2}{g_{\text{moon}}} - \frac{v_0^2}{2\cdot g_{\text{moon}}} \\ \color{red}\mathbf{h_{\text{max}} } &\color{red}\mathbf{=}& \color{red}\mathbf{ \frac{v_0^2}{2\cdot g_{\text{moon}}} }\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline h_{\text{max}} & = & \frac{v_0^2}{2\cdot g_{\text{moon}}} \qquad & | \qquad v_0 = 10\ \frac{m}{s} \qquad g_{\text{moon}} = 1.6\frac{m}{s^2} \\ h_{\text{max}} & = & \frac{ (10\ \frac{m}{s})^2}{2\cdot 1.6\frac{m}{s^2} } \\ h_{\text{max}} & = & \frac{10^2}{2\cdot 1.6} \cdot \frac{m^2}{s^2}\cdot \frac{s^2}{m}\\ h_{\text{max}} & = & \frac{100}{2\cdot 1.6} \ m\\ \mathbf{h_{\text{max}} }& \mathbf{=} & \mathbf{31.25 \ m }\\ \hline \end{array}$

The golf ball does go high 31.25 m.

The height  of the ball above the Moon's surface at any time "t"  is given by :

h(t)  = (-1.6/2)t^2  + 10t

And the max height is given by :

c - b^2 / [4a]   =

0 - 10^2/ [ 4 (-1.6/2)]  =

-100/ [ -3.2]  = 

31.25 m