## College Physics (4th Edition)

The magnitude of the change in the object's momentum is $100~kg~m/s$
It takes 1.7 seconds for the object to reach maximum height. We can find the initial vertical velocity: $v_{fy} = v_{0y}+a_y~t$ $v_{0y} = v_{fy}-a_y~t$ $v_{0y} = 0-(-9.80~m/s^2)(1.7~s)$ $v_{0y} = 16.66~m/s$ The velocity just before the object hits the ground is $-16.66~m/s$. The magnitude of the change in the vertical component of velocity is $33.32~m/s$ The horizontal component of velocity does not change while the object is in the air. Therefore, the magnitude of the object's change in velocity is $33.32~m/s$. We can find the magnitude of the change in the object's momentum: $\Delta p = m~\Delta v$ $\Delta p = (3.0~kg)(33.32~m/s)$ $\Delta p = 100~kg~m/s$ The magnitude of the change in the object's momentum is $100~kg~m/s$.