Question 3.9: A ball is thrown upward from the top of a building with an i......
A ball is thrown upward from the top of a building with an initial velocity v_{o} = 20 m/s. The building is 40 m high and the ball just misses the edge of the building roof on its way down; see Fig. 3.15 and take g = 10 m/s². Neglecting air resistance, find: (a) the time t_{1} for the ball to reach its highest point, (b) how high will it rise, (c) how long will it take to return to its starting point, (d) the velocity v_{2} of the ball at this instant, and (e) the velocity v_{3} and the total time of flight t_{3} just before the ball hits the ground.
(a) We choose upward as positive, i.e. a = −g = −10 m/s² during ascending and descending motions. Also, we choose the origin at the top of the building, i.e. y_{o} = 0, see Fig. 3.15. Since at the maximum height the ball stops momentarily, we use v_{o} = 20 m/s and v_{1} = 0 in v_{1} = v_{o} − g t_{1} to find t_{1} as follows:
0 = 20 m/s − (10 m/s²) t_{1}
t_{1}={\frac{20~{\mathrm{m/s}}}{10~{\mathrm{m/s}}^{2}}}=2~s(b) For the maximum height, we use the notation y_{1} ≡ y_{max}. To find the maximum height from the position of the thrower, we use the formula y_{max} − y_{o} = v_{o} t_{1} − \frac{1}{2} g t^{2}_{1} as follows:
y_{\mathrm{max}}=(20~\mathrm{m/s})\times(2s)-{\textstyle{\frac{1}{2}}}(10~\mathrm{m/s^{2}})\times(2s)^{2}=20~\mathrm{m}(c) When the ball returns to its starting point, the y coordinate is zero again, i.e. y_{2}=0.\;\mathrm{To}\;\mathrm{find}\;t_{2}\;\mathrm{we}\;\mathrm{use}\;y_{2}-y_{o}=v_{o}\;t_{2}-{\textstyle{\frac{1}{2}}}\;g\;t_{2}^{2} as follows (after omitting the units temporarily, since they are consistent):
0=20\;t_{2}-{\frac{1}{2}}\times10\times t_{2}^{2}This equation can be factored to give:
t_{2}[20-5 \ t_{2}]=0One solution is t_{2} = 0, which corresponds to the time that the ball starts its motion.
The other solution is t_{2} = 4 s, which is the solution we are after. Thus:
t_{2} = 4 s
(d) The value t_{2} = 4 s found in part (c) can be inserted into the formula v_{2}=v_{\mathrm{o}}-g\,t_{2} as follows:
Note that the velocity of the ball when it returns to its starting point is equal in magnitude to its initial velocity but opposite in direction. This indicates that the motion is symmetric, and generally we have:
v_{2}=-v_{o}(e) When the ball reaches the ground, its position is y_{3} = −40 m. We can insert this value in v_{3}^{2}=v_{o}^{2}-2\ g(y_{3}-y_{o}) to find v_{3} as follows:
\ v_{3}^{2}=(20~\mathrm{m/s})^{2}-2\times(10~\mathrm{m/s}^{2})[(-40~\mathrm{m})-0]=1,200\mathrm{ \ m^{2}/s^{2}\quad}Thus:
v_{3}=\pm{\sqrt{1.200{\mathrm {\ m}}^{2}/s^{2}}}=\pm34.64{\mathrm{\ m}}/sSince the ball is moving downward, we choose the negative value. Thus:
v_{3}=-34.64\;{\mathrm{m/s}}To find the total time of flight t_{3}, we use v_{3} = v_{o} − g t_{3} as follows:
t_{3}=\frac{v_{o}-v_{3}}{g}=\frac{(20\;\mathrm{m}/s)-(-34.64\;\mathrm{m}/s)}{10\;\mathrm{m}/s^{2}}=\frac{54.64\;\mathrm{m}/s}{10\;\mathrm{m}/s^{2}}=5.5\;\mathrm{s}