An object is dropped off of a 27 meter tall building.
How long is it in the air before it hits the ground?
To measure the height of a tree, a person throws a ball vertically upward, and takes 13 seconds to reach its maximum height which is exactly the height of the tree.
How tall is the tree?
One question at a time, please.
The distance a dropped ball falls in time t is
y = (g/2)t^2
In the first question, use that formula to solve for t.
In the second question, I believe your time tom reach the top of the tree should be 1.3 seconds, not 13. Otherwise the tree is unrealistically high. Recheck the problem statement.
To solve both of these problems, we can use the equations of motion for free-falling objects. The equation we will use is:
h = (1/2) * g * t^2
Where:
- h is the height (or displacement) of the object
- g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- t is the time spent in the air
For the first problem: "An object is dropped off of a 27 meter tall building. How long is it in the air before it hits the ground?"
We are given the initial height h as 27 meters, and we need to find the time t.
Rearranging the equation to solve for time:
t = √(2h/g)
Substituting the given values:
t = √(2 * 27 / 9.8)
t = √(54 / 9.8)
t = √5.51
t ≈ 2.35 seconds
Therefore, the object will be in the air for approximately 2.35 seconds before hitting the ground.
For the second problem: "To measure the height of a tree, a person throws a ball vertically upward and takes 13 seconds to reach its maximum height, which is exactly the height of the tree. How tall is the tree?"
Here, we are given the time t as 13 seconds, and we need to find the height h.
Using the same equation, but this time solving for height:
h = (1/2) * g * t^2
Substituting the given values:
h = (1/2) * 9.8 * 13^2
h = (1/2) * 9.8 * 169
h ≈ 828.9 meters
Therefore, the tree is approximately 828.9 meters tall.