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

$4.047m/s$
The required speed can be calculated as $3F=W$ $\implies 3(10t^2+300)=100(9.81)$ $\implies t=1.643s$ Now, $mv_{y1}+\Sigma \int _{t_1}^{t_2}F_ydt=mv_{y2}$ We plug in the known values to obtain: $0+\int_{1.643}^4 3(10t^2+300)dt-100(9.81)(4-1.643)=100v$ $\implies v=4.047m/s$