f a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower? Given: g = –9.8 meters/second2
Physics
torismith139
Question
f a ball dropped from a tower reaches the ground after 3.5 seconds, what is the height of the tower? Given: g = –9.8 meters/second2
2 Answer

1. User Answers SapphireMoon
Let's use the second kinematic equation:
Δy = vot + 1/2gt²
Δy = 0(3.5) + 1/2(9.8)(3.5)²
Δy = 4.9(3.5)² = 60.03 m
So the ball falls 60.03 m, meaning that's the height of the building. The negative sign just means I put the y = 0 mark at the top of the building.
Make sense? 
2. User Answers whitneytr12
The height of the tower from which a ball is dropped is 60.03 m.
The height of the tower can be calculated with the following equation:
[tex] y_{f} = y_{i} + v_{i_{y}}t  \frac{1}{2}gt^{2} [/tex]
Where:
[tex] y_{f} [/tex]: is the final height = 0 (when it reaches the ground)
[tex] y_{i} [/tex]: is the initial height =?
[tex] v_{i_{y}}[/tex]: is the initial velocity = 0 (the ball is dropped)
t: is the time = 3.5 s
g: is the acceleration due to gravity = 9.8 m/s²
Hence, the height is:
[tex] y_{i} = \frac{1}{2}9.8 m/s^{2}*(3.5 s)^{2} = 60.03 m [/tex]
Therefore, the height of the tower is 60.03 m.
You can find another example of freefall here https://brainly.com/question/9828119?referrer=searchResults
I hope it helps you!