Body trajectory energy

A body of mass 4g is shot perpendicularly from the ground and lands in 8s. What is its kinetic energy at the beginning of its trajectory, what speed does it take off, and where does the distance from the ground stop? Thank you

Final Answer:

E₁ =  3.0796 J
v =  39.24 m/s
s =  78.48 m

Step-by-step explanation:

$$\begin{gathered} m=4\text{g}\to \text{kg}=4:1000\text{kg}=0.004\text{kg} \\ t=8\text{s} \\ {t}_1=t/2=8/2=4\text{s} \\ g=9.81{\text{m/s}}^2 \\ s=\frac{1}{2}\cdot g\cdot t_1^2=\frac{1}{2}\cdot 9.81\cdot 4^2=\frac{1962}{25}=78.48\text{m} \\ {E}_1=E_2 \\ {E}_1=m\cdot g\cdot s=0.004\cdot 9.81\cdot 78.48=3.0796\text{J} \end{gathered}$$

$v=g\cdot t_1=9.81\cdot 4=39.24\text{m/s}$

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