Respuesta :
Answer:
(a) [tex]y_{max}=0.423m[/tex]
(b) [tex]\alpha =64.3^{o}[/tex]
Explanation:
Given data
[tex]v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle[/tex]
Solution
For Part (a)
As the velocity component in direction of y is given by:
[tex]v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s[/tex]
The maximum displacement is given by:
[tex]v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m[/tex]
For Part (b)
To reach y=46cm =0.46m apply:
[tex]0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}[/tex]