Here is the math problem:
The height of an object in free fall, or otherwise thrown into the air near the earth's surface, with initial vertical velocity v0 in feet per seconds, and from the initial height h0 in feet, at any given time t in seconds is given by the equation:
h(t)= -16t^2 + v0t + h0
At 1821 feet tall, the CN Tower in Toronto, Canada is the world's tallest...etc. Suppose an object is dropped from the top of the tower (just released not pushed) .
1. The equation above find h0 and v0
2. The height of the object 2.3 seconds after being released.
Please help, even if you could show me how to proper start this problem I would be thankful! :]
An object moving vertically is at the given heights at the specified times. Find the position equation
s =
1
2
at2 + v0t + s0
for the object.
At
t = 1 second, s = 146 feet
At
t = 2 seconds, s = 98 feet
At
t = 3 seconds, s = 18 feet
To find h0 and v0, we can use the information given about the CN Tower. Since the object is dropped from the top of the tower, its initial height h0 would be equal to the height of the CN Tower, which is 1821 feet.
Now let's focus on finding v0. When the object is dropped, its initial velocity v0 is zero because it is just released and not pushed. So v0 = 0.
Now let's move on to the second part of the problem, finding the height of the object 2.3 seconds after being released.
Using the equation h(t) = -16t^2 + v0t + h0, we can substitute the values we know to find the height at any given time.
In this case, we are interested in the height after 2.3 seconds, so let's substitute t = 2.3 into the equation:
h(t) = -16(2.3)^2 + 0(2.3) + 1821
Simplifying the equation, we get:
h(t) = -16(5.29) + 1821
h(t) = -84.64 + 1821
h(t) ≈ 1736.36 feet
Therefore, the height of the object 2.3 seconds after being released is approximately 1736.36 feet.