Question 3.7: A ball is dropped from a tall building, as shown in Fig. 3.1......
A ball is dropped from a tall building, as shown in Fig. 3.13. Choose the positive y to be downward with its origin at the top of the building, i.e. y_{o} = 0. Find the following for the ball’s motion: (a) its acceleration, (b) the distance it falls in 2 s, (c) its velocity after falling 15 m, (d) the time it takes to fall 25 m, and (e) the time it takes to reach a velocity of 29.4 m/s.
(a) Since the positive y is downward, then the ball’s acceleration is positive (downward) and will be given by a = g = 9.8 m/s². Also, the ball’s velocity will be always positive.
(b) We are given v_{o} = 0, y_{o} = 0, a = g = 9.8 m/s², and t = 2 s. To find y, we use y − y_{o} = v_{o} t + \frac{1}{2} g t² as follows:
(c) We are given v_{o} = 0, y_{o} = 0, a = g = 9.8 m/s², and y = 15 m. To find v, we use v² = v²_{o} + 2 g(y − y_{o}) as follows:
v^{2}=0+2\times(9.8\;\mathrm{m/s}^{2})\times15\;\mathrm{m}\ \ \Rightarrow\ \ \ v=\pm\sqrt{294}\;\mathrm{m/s}\ \ \Rightarrow\ \ \ v=17.2\;\mathrm{m/s}(d) We are given v_{o} = 0, y_{o} = 0, y = 25 m, and a = g = 9.8 m/s². To findt, we use y − y_{o} = v_{o} t + \frac{1}{2} g t² as follows:
25=0+\frac{1}2\left(9.8\mathrm{~m}/\mathrm{s}^{2}\right)\times t^{2}\quad\Rightarrow\quad t=\pm\sqrt{5.1}\ \mathrm{s}\quad\Rightarrow\quad t=2.3\ \mathrm{s}(e) We are given v_{o} = 0, v = 29.4 m/s, and a = g = 9.8 m/s². To find t, we use v = v_{o} + g t as follows:
t={\frac{v-v_{o}}{g}}={\frac{29.4\,{\mathrm{m}}/s-0\,{\mathrm{m}}/s}{9.8\,{\mathrm{m}}/s^{2}}}=3\,s