Answer:
a) 9.8 m/s
b) 4.9 m
Explanation:
This problem is a good example of Vertical motion, where the main equations for this situation are:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
[tex]V^{2}={V_{o}}^{2}-2gy[/tex] (2)
Where:
[tex]y[/tex] is the height of the ball at a given time
[tex]y_{o}=0m[/tex] is the initial height of the ball (assuming the hand of the thrower the origin of the system)
[tex]V_{o}[/tex] is the initial velocity of the ball
[tex]V[/tex] is the final velocity of the ball
[tex]t=2s[/tex] is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due to gravity
Knowing this, let's begin with the answers:
In order to find the initial velocity [tex]V_{o}[/tex] of the ball, we will use equation (1) and [tex]t=2s[/tex], taking into account that [tex]y=0 m[/tex] and [tex]y_{o}=0m[/tex] at this given time:
[tex]0=0+V_{o}t-\frac{1}{2}gt^{2}[/tex] (3)
Isolating [tex]V_{o}[/tex]:
[tex]V_{o}=\frac{1}{2}gt[/tex] (4)
[tex]V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)[/tex] (5)
Then:
[tex]V_{o}=9.8 m/s[/tex] (6)
In this part, we will use equation (2), knowing the value of the height is maximum when [tex]V=0[/tex]. So, we will name this height as [tex]y_{max}[/tex]:
[tex]0={V_{o}}^{2}-2gy_{max}[/tex] (7)
Isolating [tex]y_{max}[/tex]:
[tex]y_{max}=\frac{{V_{o}}^{2}}{2g}[/tex] (8)
[tex]y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}[/tex] (9)
Finally:
[tex]y_{max}=4.9 m[/tex]
