What is the angle of elevation θ that is required for a hit ball to just clear the center field fence when it is hit 2 ft above the ground? Show all of your work and round your answer to the nearest hundredth of a degree.
Distance from Home plate to the center field fence: 425 ft
Height of the center field fence: 8 ft
The y component of the hit will be Velocity * sin θ
This has two unkonowns....were you given the initial velocity of the hit?
the higher the initial velocity, the smaller the angle theta required
y = yo + vo t - 1/2 a t^2 this will need the y velocity to calculate the time to the point it is 8 ft off of the ground
then the time found will have to be plugged into velocity * cos θ t to calculate the fence clearance distance 425 ft
Hey you must be in geometry for k12, im doing the same project.
For this, you have to use inverse tangent to find theta. The terms were opposite, adjacent. You take inverse tangent because you are looking for the angle.
Tan-1(8/425) = 1.078 BUT!!!!
It is not 8 ft BECAUSE you probably just took the fence height and thats it. You can't do that because then the fence height itself is taking the 2 ft from the ground to the bat which doesn't count towards the angle. You take 8-2 to get the fence height from the BAT to the FENCE TOP.So, that would be 6ft. Also, the sentence did say "hit 2 ft above the ground" which means you wouldn't use those 2 ft for the height
Tan-1(6/425) = 0.8088
As you can see, that angle elevation would count for the baseball field that has an angle elevation of less than 0.86 degrees, this is where this question came into play, basically answering that.
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