Body trajectory energy

A body of mass 4 g is shot perpendicularly from the ground and lands in 8 s. 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?

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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