# [SOLVED]Lunar Gravity velocity on impact

#### karush

##### Well-known member
Lunar Gravity, On the moon, the acceleration due to gravity is $-1.6m/s$ . A stone dropped from a cliff on the moon and hits the surface of the moon $20s$ later.
How far did it fall?
What is the velocity at impact.?

well here I presume since the rock is just dropped that $v_0=1.6m/s$ but the time is known t=20s and

$f(t)=-1.6t^2+1.6t+S_0$

$f'(t)=-3.2t+1.6 but I don't think this is set up right. Ans 320m; -32m/s #### MarkFL ##### Administrator Staff member Just a trivial matter perhaps, but the units of acceleration are distance per time squared. Now, recall the kinematic equation for displacement given constant acceleration is:$\displaystyle s(t)=\frac{1}{2}at^2+v_0t+s_0$Since the rock is dropped, its initial velocity is zero. Its final height is zero, and it fell for 20 seconds. Hence, we may state:$\displaystyle f(t)=-0.8t^2+s_0$Solve for$\displaystyle s_0$given:$\displaystyle f(20)=0$To find the velocity at impact, evaluate:$\displaystyle f'(20)$#### CaptainBlack ##### Well-known member Lunar Gravity, On the moon, the acceleration due to gravity is$-1.6m/s$. A stone dropped from a cliff on the moon and hits the surface of the moon$20s$later. How far did it fall? What is the velocity at impact.? well here I presume since the rock is just dropped that$v_0=1.6m/s$but the time is known t=20s and$f(t)=-1.6t^2+1.6t+S_0f'(t)=-3.2t+1.6

but I don't think this is set up right.

Ans 320m; -32m/s
1. The units of the acceleration due to gravity are $$m/s^2$$

2. Dropped usually means that $$v_0=0$$

3. $$\displaystyle s(t)=\frac{g\; t^2}{2}+v_0 t +s_0 = - \frac{1.6\; t^2}{2}+s_0$$

When the rock hits the surface $$s(t)=0$$, so ...

4. $$\displaystyle v(t)=g\;t+v_0 =$$ ...

5. Since this is in calculus you need the supporting calculus: The equations in (3) and (4) above are obtained by integrating $$\ddot{s}= g$$ and using $$v(t)=\dot{s}(t)$$ with initial conditions $$\dot{s}(0)=v(0)=v_0$$ and $$s(0)=s_0$$.

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#### karush

##### Well-known member
thanks much everyone, that was a great help $0=-0.8(20)^2+S_o\ \ S_o=320m$

$f'(20)=-1.6(20) =-32m/s$

will be back with more, seem to slip and slide with this stuff.