Answer:
a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) The total distance traveled by the baseball was 108.7 m.
Explanation:
a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:
[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex] y_{f}[/tex]: is the final height =?
[tex] y_{0}[/tex]: is the initial height = 1 m
[tex]v_{0_{y}[/tex]: is the initial vertical velocity = v₀sin(45)
v₀: is the initial velocity = 32.5 m/s
g: is the gravity = 9.81 m/s²
t: is the time
First, we need to find the time by using the following equation:
[tex]t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s[/tex]
Now, the height is:
[tex]y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m[/tex]
Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.
b) To find the distance traveled by the baseball first we need to find the time of flight:
[tex] y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} [/tex]
[tex]0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}[/tex]
[tex]1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0[/tex]
By solving the above quadratic equation we have:
t = 4.73 s
Finally, with that time we can find the distance traveled by the baseball:
[tex] x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m [/tex]
Hence, the total distance traveled by the baseball was 108.7 m.
I hope it helps you!
