Free fall

The free fall body has gone 10m in the last 0.5s. Find the body speed at the moment of impact.

Result

v =  22.5 m/s

Solution:

t=0.5 s s=10 m  g=10 m/s2    s=12⋅ g⋅ t12−12⋅ g⋅ (t1−t)2 2⋅ s/g=t12−(t1−t)2 2⋅ s/g=t12−t12+2⋅ t⋅ t1−t2 2⋅ s/g+t2=2⋅ t⋅ t1  t1=t2+2⋅ s/g2⋅ t=0.52+2⋅ 10/102⋅ 0.5=94=2.25 s  v=g⋅ t1=10⋅ 2.25=452=22.5 m/st=0.5 \ \text{s} \ \\ s=10 \ \text{m} \ \\ \ \\ g=10 \ \text{m/s}^2 \ \\ \ \\ \ \\ \ \\ s=\dfrac{ 1 }{ 2 } \cdot \ g \cdot \ t_{1}^2 - \dfrac{ 1 }{ 2 } \cdot \ g \cdot \ (t_{1}-t)^2 \ \\ 2 \cdot \ s/g=t_{1}^2 - (t_{1}-t)^2 \ \\ 2 \cdot \ s/g=t_{1}^2 - t_{1}^2 + 2 \cdot \ t \cdot \ t_{1} - t^2 \ \\ 2 \cdot \ s/g + t^2=2 \cdot \ t \cdot \ t_{1} \ \\ \ \\ t_{1}=\dfrac{ t^2 + 2 \cdot \ s/g }{ 2 \cdot \ t }=\dfrac{ 0.5^2 + 2 \cdot \ 10/10 }{ 2 \cdot \ 0.5 }=\dfrac{ 9 }{ 4 }=2.25 \ \text{s} \ \\ \ \\ v=g \cdot \ t_{1}=10 \cdot \ 2.25=\dfrac{ 45 }{ 2 }=22.5 \ \text{m/s}

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