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# Straight throw, determination of the initial speed

A lorry fell into a shaft and left traces on the wall at depth H under the crash site. At what speed did the lorry go into the shaft? (The air resistance can be neglected.)

Given: H = 20 m, b = 4,5 m, g = 9,81 m/s²

### Solution

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The acceleration is

$\tag{1} \ddot{x} = 0$

$\tag{2} \ddot{z} = -g$

The functions in the x and z directions for speed and position are obtained through double integration.

$\tag{3} \dot{x} = c_1$

$\tag{4} x(t) = c_1t+c_3$

$\tag{5} \dot{z}(t) = -gt + c_2$

$\tag{6} z(t) = -\frac{1}{2}gt^2 + c_2t + c_4$

The speed at time t = 0

$\tag{7} \dot{\vec{x}}(t = 0) = v_0$

$\tag{8} \dot{\vec{z}}(t = 0) = 0$

which means that c1 and c2 can be calculated.

$\require{cancel}$

$\tag{9} v_0 = c_1$

$\tag{10} c_1 = v_0$

$\tag{11} 0 = \bcancel{-gt} + c_2$

$\tag{12} c_2 = 0$

From the initial conditions for the position at time t = 0, c3 and c4 can be calculated.

$\require{cancel}$

$\tag{13} \vec{x}(t = 0) = 0$

$\tag{14} 0 = \bcancel{v_0t}+c_3$

$\tag{15} c_3 = 0$

$\tag{16} \vec{z}(t = 0) = 0$

$\tag{17} 0 = \bcancel{-\frac{1}{2}gt^2} + c_4$

$\tag{18} c_4 = 0$

The as yet unknown point in time of the impact is now referred to as T. The position of the cart at time T is:

$\require{cancel}$

$\tag{19} \vec{x}(t = T) = b$

$\tag{20} \vec{z}(t = T) = -H$

From which both the time T and the initial speed v0 can be calculated in the following.

$\require{cancel}$

$\tag{21} b = v_0T$

$\tag{22} -H = -\frac{1}{2}gT^2$

$\tag{23} T = \sqrt{\frac{2H}{g}}$

$\tag{24} v_0 = \frac{b}{\sqrt{\frac{2H}{g}}}$

$\tag{25} v_0 = \frac{ 4.5 m}{ \sqrt{ \frac{ 40 m}{ 9.81 \frac{ m }{ s^2 }}}} = 2.23 \frac{ m }{ s }$