Answer:
V0=27.4 m/s; t=0.8 s
Explanation:
Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:
[tex]y_f=v_0t-0.5*g*t^2[/tex] We solve for initial speed
[tex]v_0=\frac{y_f+0.5gt^2}{t}=\frac{37+4.9*2.3^2}{2.3}=27.4m/s[/tex]
Now, using the same expression we estimated time to first reach 18.5 m :
[tex]18.5=27.4t-4.9t^2[/tex] Second order equation with solutions
t1=0.8 s and t2=4.8 s
The first time corresponds to the first reach.
