Answer:
[tex](a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s[/tex]
Explanation:
Given data
[tex]v_{o}=7.14m/s\\\alpha =36.5^o[/tex]
For Part (a)
Starting with the -ve acceleration of the body (opposite to the gravitational force)
[tex]a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2[/tex]
Using equation of motion
[tex]v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m[/tex]
For Part (b)
Using the result in Part (a) we can substitute in other equation of motion to get time t:
[tex]x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s[/tex]
For Part (c)
At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)
[tex]a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s[/tex]
