Respuesta :
Answer:
d). The value of y should be -32m
Vx=0.92 m/s
Explanation:
Using equation of motion uniform to y motion
[tex]x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\[/tex]
So to find t that is the same time for all the motion
[tex]t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s[/tex]
The value of Xf=-3.2m because the g is negative from the axis
Now in the axis 'x' to find Vx
[tex]Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}[/tex]
Answer:
D The value of y should be –3.2 m.
Explanation:
just got it right