Answers:
a) 5.51 s
b) 37.193 m
c) 1.377 s
Explanation:
This situation is described by Vertical motion, where the main equation is:
[tex]y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0[/tex] is the inal height of the ball
[tex]y_{o}=[/tex] is the initial height of the ball
[tex]V_{o}=27 m/s[/tex] is the initial velocity of the ball
[tex]t[/tex] is the time at a certain height
[tex]g=-9.8m/s^{2}[/tex] is the acceleration due to gravity (always directed donwards)
Knowing this, let's begin with the answers:
In this part, we have to calculate the total time [tex]t_{T}[\tex] the ball is in the air. So, we have to find the time from (1) and taking into account the conditions given above:
[tex]0=V_{o}t_{T}+\frac{1}{2}g{t_{T}}^{2}[/tex] (2)
Isolating [tex]t_{T}[/tex]:
[tex]t_{T}=\frac{-2V_{o}}{g}[/tex] (3)
[tex]t_{T}=\frac{-2(27 m/s)}{-9.8m/s^{2}}[/tex] (4)
[tex]t_{T}=5.51 s[/tex] (5)
The ball reaches its maximum height [tex]y_{max}[/tex] when the time [tex]t[/tex] is half the total time [tex]t_{T}[/tex]. Just in the moment the ball reaches the half of its parabolic path and begins falling down:
[tex]t=\frac{t_{T}}{2}=\frac{5.51 s}{2}[/tex]
[tex]t=2.75 s[/tex] (6)
Now using this time in equation (1) but renaming [tex]y[/tex] as [tex]y_{max}[/tex]:
[tex]y_{max}=V_{o}t+\frac{1}{2}gt^{2}[/tex] (7)
[tex]y_{max}=(27 m/s)(2.75 s)+\frac{1}{2}(-9.8m/s^{2})(2.75 s)^{2}[/tex] (8)
[tex]y_{max}=37.193 m[/tex] (9)
At this point (considering we are dealing with constant acceleration and the movement is in th vertical direction) we can use the following equation:
[tex]V=V_{o}+gt[/tex] (10)
Where:
[tex]V=\frac{V_{o}}{2}[/tex] is the velocity we want to find, when it is half of the ball's initial velocity
Rewritting (10):
[tex]\frac{V_{o}}{2}=V_{o}+gt[/tex] (11)
Isolating [tex]t[/tex]:
[tex]t=\frac{-V_{o}}{2g}[/tex] (12)
[tex]t=\frac{-27 m/s}{2(9.8 m/s^{2})}[/tex] (13)
Finally:
[tex]t=1.377 s[/tex] This is the first time where the ball has half of its initial velocity
