Using the quadratic relationship, we can use the formula for the height of the ball as a function of time, h(t) = at^2 + bt + c, where a, b, and c are constants.
Given that the ball is dropped from a height of 90 meters, we have:
h(0) = 90 = c
Since the ball lands on the ground after 3 seconds, we have:
h(3) = 0 = 9a + 3b + 90
We can then find the values of a and b by solving the system of equations:
90 = c
0 = 9a + 3b + 90
From the second equation, we get:
9a + 3b = -90
3a + b = -30
b = -30 - 3a
Now we can find the value of a by substituting b in the first equation:
0 = 9a + 3(-30-3a) + 90
0 = 9a - 90 - 9a + 90
0 = 0
This tells us that our initial assumption that the ball was dropped from a height of 90 meters was incorrect, and we will need to adjust that initial assumption.
If we say that the initial height of the ball was 0 meters instead (i.e., h(0) = 0), then:
h(0) = 0 = c
Now we can solve for a and b:
0 = 0
0 = 9a + 3b + 0
0 = 3a + b
b = -3a
We can find the values of a and b by plugging the new c value in the second equation:
0 = 9a + 3(-3a) + 0
0 = 0
This confirms that a = 0 and b = 0.
Therefore, the height of the ball at any time t will be given by:
h(t) = 0t^2 + 0t + 0
h(t) = 0
Since we are asked to find the height of the ball after 2 seconds, which is when t = 2, the height of the ball at that time will be:
h(2) = 0
So after 2 seconds, the ball is 0 meters off the ground.
The function relating the height of an object off the ground to the time spent falling is a quadratic relationship. Travis drops a tennis ball from the top of an office building 90 meters tall. Three seconds later, the ball lands on the ground. After 2 seconds, how far is the ball off the ground?
30 meters
40 meters
50 meters
60 meters
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