Answer:
Assumption: the air resistance on this ball is negligible. Take [tex]g = 10\; \rm m \cdot s^{-2}[/tex].
a. The momentum of the ball would be approximately [tex]60\;\rm kg \cdot m \cdot s^{-1}[/tex] two seconds after it is tossed into the air.
b. The momentum of the ball would be approximately [tex]\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)[/tex] three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.
Explanation:
The momentum [tex]p[/tex] of an object is equal its mass [tex]m[/tex] times its velocity [tex]v[/tex]. That is: [tex]\vec{p} = m \cdot \vec{v}[/tex].
Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant [tex]g \approx -10\; \rm m \cdot s^{-2}[/tex]. In other words, its velocity would become approximately [tex]10\; \rm m \cdot s^{-1}[/tex] more negative every second.
The initial velocity of the ball is [tex]60\; \rm m \cdot s^{-1}[/tex]. After two seconds, its velocity would have become [tex]60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}[/tex]. The momentum of the ball at that time would be around [tex]p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}[/tex].
When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant [tex]g \approx -10\; \rm m \cdot s^{-2}[/tex]. That's how the ball's velocity becomes negative.
After three more seconds, the velocity of the ball would be [tex]0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}[/tex]. Accordingly, the ball's momentum at that moment would be [tex]p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)[/tex].
