Question 3.8: A boy throws a ball upwards, giving it an initial speed vo =......
A boy throws a ball upwards, giving it an initial speed v_{o} = 15 m/s. Neglect air resistance. (a) How long does the ball take to return to the boy’s hand? (b) What will be its velocity then?
(a) We choose the positive y upward with its origin at the boy’s hand, i.e. y_{o} = 0, see Fig. 3.14. Then, the ball’s acceleration is negative (downward) during the ascending and descending motions, i.e. a = −g = −9.8 m/s². When the ball returns to the boy’s hand its position y is zero. Since v_{o} = 15 m/s, y_{o} = 0, y = 0, and a = −g, then we can find t from y − y_{o} = v_{o} t − \frac{1}{2} g t^{2} as follows:
0=(15\,{\mathrm{m/s}})\;t-{\frac{1}{2}}\,({{9}}.{\mathrm{8}}\,{\mathrm{m/s}}^{2})\times t^{2}\quad\Rightarrow\quad t={\frac{2\times(15{\mathrm{m/s}})}{9.8\,{\mathrm{m/s}}^{2}}}=3.1\,{\mathrm{s}}(b) We are given v_{o} = 15 m/s, y_{o} = 0, y = 0, and a = −g = −9.8 m/s². To find v, we use v² = v²_{o} − 2 g(y − y_{o}) as follows:
v^{2}=v_{o}^{2}-0\;\;\;\Rightarrow\;\;\;v=\pm\sqrt{v_{o}^{2}}=\pm v_{o}=\pm15\;\mathrm{m/s}We should select the negative sign, because the ball is moving downward just before returning to the boy’s hand, i.e. v = −15 m/s.
