Rocket start

The body launched vertically up returns to the start site in 6 seconds. What height did it have?

Result

h =  45 m

Solution:

g=10 m/s2 t=6 s  t1=t/2=6/2=3 s  h=12 g t12=12 10 32=45 mg=10 \ \text{m/s}^2 \ \\ t=6 \ \text{s} \ \\ \ \\ t_{1}=t/2=6/2=3 \ \text{s} \ \\ \ \\ h=\dfrac{ 1 }{ 2 } \cdot \ g \cdot \ t_{1}^2=\dfrac{ 1 }{ 2 } \cdot \ 10 \cdot \ 3^2=45 \ \text{m}



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