## Engineering Mechanics: Statics & Dynamics (14th Edition)

$h=1.57m$
The required height can be determined as follows: We know that: $e=\frac{v_{g2}-v_{By}^{\prime}}{v_{By}-v_{g1}}$ We plug in the known values to obtain: $0.8=\frac{0-v_{By}^{\prime}}{-6.946-0}$ This simplifies to: $v_{By}^{\prime}=5.556m/s\uparrow$ From the equation of kinematics $v_{Cy}^2=v_{By}^{\prime 2}+2a_c(S_{Cy}-S_{By})$ We plug in the known values to obtain: $0=(5.556)^2+2(-9.81)h$ This simplifies to: $h=1.57m$