Question 4.4: A ball thrown from the top of a building has an initial spee......
A ball thrown from the top of a building has an initial speed of 20 m/s at an angle of 30° above the horizontal. The building is 40 m high and the ball takes time t′ before hitting the ground, see the Fig. 4.12. Take g = 10 m/s². (a) Find 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 the level of the thrower? (d) Find the time of flight t′. (e) What is the horizontal distance covered by the ball during this time? (f) What is the velocity of the ball before striking the ground?
(a) Since y is up, then a = −g = −10 m/s² during ascending and descending motions. Also, since the origin is at the top of the building, then y_{o} = 0. The initial components of the velocity are:
v_{x o}=v_{o}\cos\theta_{o}=(20\,{\mathrm{m/s}})\,\left(\cos30^{o}\right)=17.32\,{\mathrm{m/s}},
v_{y o}=v_{o}\sin\theta_{o}=(20\,{\mathrm{m/s}})\,\left(\sin30^{o}\right)=10\,{\mathrm{m/s}}
Since at the maximum height the ball stops momentarily, we use v_{y0} = 10 m/s and v_{y} = 0 in v_{y} = v_{yo} − g t to find t_{1} as follows:
0=10\;\mathrm{m/s}-(10\;\mathrm{m/s}^{2})\;t_{1}\quad\Rightarrow\quad t_{1}={\frac{10\;\mathrm{m/s}}{10\;\mathrm{m/s}^{2}}}=1\;s(b) To find the maximum height H from the position of the thrower, we use t_{1}=1 \ s \ and \ y =v_{y o}t-{\frac{1}{2}}g t^{2}, or Eq. 4.24, as follows:
H = \frac{v^{2}_{o} sin^{2} θ_{o}}{ 2g} , 0 ≤ \theta_{o} ≤ π/2 (4.24)
H=(10\;\mathrm{m/s})\times(1\,\mathrm{s})-\frac{1}{2}(10\;\mathrm{m/s}^{2})\times(1\,\mathrm{s})^{2}=5\;\mathrm{m}Thus, the maximum height of the ball from the ground is 45 m.
(c) When the ball returns to the level of the thrower, the y coordinate is zero again, i.e. y = 0, and t = T. To find the time T the ball takes to reach this location, we use y = v_{yo}t − \frac{1}{2} gt^{2} as follows (after omitting the units temporarily since they are consistent):
We can also use T = 2t_{1} = 2 s for the symmetric part of the path.
(d) To find t′, we can use y = v_{yo}t − \frac{1}{2} gt^{2} with y = −40 m and v_{yo} = 10 m/s so that (after omitting the units):
Solving this quadratic equation yields:
t^{\prime}={\frac{10\pm{\sqrt{(-10)^{2}-4\times5\times(-40)}}}{2\times5}}={\frac{10\pm30}{10}}\quad\Rightarrow\ \ t^{\prime} =\left\{\begin{matrix} +4 \ s\\-2 \ s\end{matrix}\right.We reject the negative time and take only the positive root, i.e. t′ = 4 s.
(e) The horizontal distance x covered by the ball at t′ = 4 s is:
x = (v_{o} cos θ_{o}) t^\prime = v_{xo} t^\prime
= (17.32 m/s)(4 s) = 69.28 m
(f) The vertical component of the ball’s velocity at t′ = 4 s is given by:
v_{y} = v_{yo} − gt′ = 10 m/s − (10 m/s²) (4 s) = −30 m/s
The negative sign indicates that the vertical component is directed downwards.
Since v_{x} = v_{xo} = 17.32 m/s, the required speed would be:
The direction of \vec{v} at t′ = 4 s is indicated in the Fig. 4.12 by the angle θ. Thus, according to this figure we have:
\theta=\tan^{-1}\left({\frac{|v_{y}|}{v_{x}}}\right)=\tan^{-1}\left({\frac{30{\mathrm{~m/s}}}{17.32{\mathrm{~m/s}}}}\right)=\tan^{-1}(1.73)=60^{\circ}