(Solved): HELP EXAMPLE 2-12 Ball thrown upward. A person throws a ball upacard into the air with an initial ve ...

EXAMPLE 2-12 Ball thrown upward. A person throws a ball upacard into the air with an initial velocity of $$15.0\mathrm{m}/\mathrm{s}$$. Calculate how high it gocs. Ignore air resistance. APPPOACH We are not concerned here with the throwing action, but only with the motion of the ball after it leaves the thrower's hand (Fig. 2-23) and until it comes back to the hand again. Let us choose $$y$$ to be positive in the upward direction and negative in the downward direction. (This is a different convention from that used in Examples 2-10 and 2-11, and so illustrates our options.) The acceleration due to gravity is downward and so will have a negative sign, $$a=?g=?9.80\mathrm{m}/{\mathrm{s}}^2$$. As the ball rises, its speed decreases until it reaches the highest point (B in Fig. $$2?23$$ ), where its speed is zero for an instant: then it descends, with increasing speed. Ball thrown upward, II. In Fig. 2-23, Example 2-12, how long is the ball in the air before it comes back to the hand? APPROACH We need to choose a time interval to calculate how long the ball is in the air before it returns to the hand. We could do this calculation in two parts by first determining the time required for the ball to reach its highest point, and then determining the time it takes to fall back down. However, it is simpler to consider the time interval for the entire motion from $$A$$ to $$B$$ to $$C$$ (Fig. 2-23) in one step and use Eq. 2-11b. We can do this because $$y$$ is position or displacement, and not the total distance traveled. Thus, at both points $$A$$ and $$C,y=0$$. An object thrown into the air leaves the thrower's hand at $$A$$, reaches its mavimum beight at $$\mathrm{B}$$, and retarno to the original position at $$\mathrm{C}$$ Examples 2-12,2-13,2-14, and 2-15.